Answer:
Explanation:
We shall represent the velocity of cruise ship and coast guard petrol boat in vector form .
velocity of cruise ship
Vcs = - 2.5 j
Vpb = - 4.8 cos 19 i + 4.8 sin 19 j = - 4.54 i + 1.56 j
velocity of the cruise ship relative to the patrol boat
= Vcs - Vpb
= - 2.5 j - ( - 4.54 i + 1.56 j )
= - 2.5 j + 4.54 i - 1.56 j
= 2.04 i - 1.56 j .
x-component of the velocity of the cruise ship relative to the patrol boat
= 2.04 m /s
y-component of the velocity of the cruise ship relative to the patrol boat
= - 1.56 m /s .
Emily takes a trip, driving with a constant velocity of 85.5 km/h to the north except for a 22 min rest stop. If Emily's average velocity is 68.8 km/h to the north, how long does the trip take? Show your work
Answer:
1½ hours
Explanation:
Let the amount of time she drove at 85.5 km/hr be t, then
The distance that she rode would be 85.5 * t
This is means Emily has traveled the same distance at 68.8 km/hr without any stoppage whatsoever. Thus, we can create an equation as such
85.5 * t = 68.8 (t + 22/60), the 22/60 being, converting 22 minutes into hours. We proceed further as
85.5t = 68.8t + 1513.6/60, multiplying all sides by 60, we have
5130t = 4128t + 1513.6, now we collect like terms
5130t - 4128t = 1513.6
1002t = 1513.6
t = 1513.6/1002
t = 1.51 hrs, or in minutes, 1.51 * 60
t = 90.63 minutes. 1 hour and 36 minutes
So the total time the trip took, 22 minutes in addition, or say, 1 hour 58 minutes.
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853 m/s2. What was the car’s velocity at point “A”?
Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the initial and final velocities, respecitvely; [tex]a[/tex] is the acceleration; and [tex]\Delta x[/tex] is the change in position.
So we have
[tex]\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)[/tex]
[tex]\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}[/tex]
(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)
The morning after a massive snowstorm, Michaela gets into her car to drive to work. The storm caused her windows to freeze, so she first needs to defrost the car. While the engine is running, she checks the thermometer. It shows the air inside of her car has a temperature of 0 °C. Does this mean the air inside of her car has no kinetic energy? Explain your answer.
Answer:
Hope it helps
A Brainliest please
What physical property of Earth gives rise to the seasons?
Answer:
Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.
Explanation:
John is gliding along on his scooter at a speed of 2.8 m/s when
suddenly he collides with Sam who is walking around the corner. If
John is brought to rest in 0.15 s, what is his acceleration?
Answer: His acceleration is -18.66m/s^2
Explanation:
Ok, the initial speed is 2.8m/s. (we can define the initial direction as the positive direction).
And he wants to stop, so he must accelerate in the opposite direction as the initial movement, then we would have:
a(t) = -A.
So we have a constant, and negative acceleration.
Now, if we want to find the velocity we must integrate over time, and we will get:
v(t) = -A*t + V0
where V0 is the initial velocity, we know that it is 2.8m/s, and t is the time in seconds.
Then the velocity is:
v(t) = -A*t + 2.8m/s.
Now we know that John is brought to rest in 0.15 seconds after he starts slowing down, this means that at t = 0.15 seconds, his velocity is equal to zero.
v(0.15s) = 0m/s = -A*0.15s + 2.8m/s
2.8m/s = A*0.15s
2.8m/s/0.15s = 18.66m/s^2 = A.
So his acceleration is -A, then we have that:
His acceleration is -18.66m/s^2
An ionized oxygen molecule (O2+) at point A has charge +e and moves at 1.24 ✕ 103 m/s in the positive x-direction. A constant electric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.766 mm past A on the x-axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of an oxygen molecule is 5.31 ✕ 10−26 kg and the fundamental charge is e = 1.60 ✕ 10−19 C.)
Answer:
[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]
[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]
Explanation:
Given that:
The charge on the ionized oxygen molecule = +e
The speed of the ionized oxygen molecule with this charge is 1.24 × 10³ m/s
distance travelled by the particle before rest is d = 0.766 m
According to the third equation of motion.
[tex]v^2 = u^2 +2as[/tex]
[tex]v^2 = u^2 +2(\dfrac{-eE}{m}) s[/tex]
[tex]0^2= u^2 +2(\dfrac{-eE}{m}) s[/tex]
[tex]E = \dfrac{mu^2}{2e* \ s}[/tex]
[tex]E = \dfrac{5.31 *10^{-26}* (1.24*10^3)^2}{2*1.6*10^{-19}*0.766*10^{-3}}[/tex]
[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]
Thus, the electric field shows to be in the negative x-direction.
The potential difference between point A and B now is:
[tex]\Delta V = E.d \\ \\ V_A - V_B = 3.33 \times 10^2 \times 0.766 \times 10^{-3}[/tex]
[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]
Part A
Suppose that you want to construct a line with slope m=3 that passes through the point (2,1). You would begin by setting up the equation
y=3x+b.
If you plug in the coordinates for any point on that line, the two sides of the equation will be equal. Once you've done this, you can solve for b. What is the value of b?
Express your answer as an integer.
Part B
Suppose that you want to find the equation for a line that passes through the two points (0,3) and (4,9). What is the slope of this line?
Express your answer numerically.
Answer:
a) b = -5
b) slope = 3/2
Explanation:
a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.
Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:
1 = 3(2) + b
1 = 6 + b
b = 1 - 6
b = -5
b) The equation of a line passing through two points ([tex]x_1,y_1[/tex]) and [tex]x_2,y_2[/tex] is given by:
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
The equation of the line passing through the two points (0,3) and (4,9) is:
[tex]y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3[/tex]
Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2
Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. When the container is full, what is the pressure that the water exerts on the sides of the container at the bottom of the container above the atmospheric pressure? b. When the container is full, what is the pressure that the water exerts on the sides of the container halfway down from the top above the atmospheric pressure?
Answer:
a) 105935.7 Pa
b) 103630.35 Pa
Explanation:
The volume of the container = 0.025 m^3
The radius of the container = 13 cm = 0.13 m
We have to find the height of the tank
From the equation for finding the volume of the cylinder,
V = [tex]\pi r^2h[/tex]
where
V is the volume of the cylinder
h is the height of the cylinder
substituting values, we have
0.025 = 3.142 x [tex]0.13^2[/tex] x h
0.025 = 0.0531h
h = 0.025/0.0531 = 0.47 m
Pressure at the bottom of the tank P = ρgh
where
ρ is the density of water = 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
h is the depth of water which is equal to the height of the tank
substituting values, we have
P = 1000 x 9.81 x 0.47 = 4610.7 Pa
atmospheric pressure = 101325 Pa
therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = 105935.7 Pa
b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m
pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa
This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = 103630.35 Pa
(a) The pressure exerted by water on the sides of the container at the bottom is 4606 Pa.
(b) The pressure exerted by water on the sides of the container halfway down from the top is of 2303 Pa.
Given data:
The volume of cylindrical container is, [tex]V = 0.025 \;\rm m^{3}[/tex].
The radius of container is, r = 13 cm = 0.13 m.
(a)
When the container is full, the pressure that the water exerts on the sides of the container is given as,
[tex]P = \rho gh[/tex]
Here, [tex]\rho[/tex] is the density of water, g is the gravitational acceleration and h is the height of container and its value is obtained as,
[tex]V = \pi r^{2}h\\\\0.025= \pi \times (0.13)^{2} \times h\\\\h = 0.47 \;\rm m[/tex]
Then pressure is,
[tex]P = 1000 \times 9.8 \times 0.47\\\\P=4606 \;\rm Pa[/tex]
Thus, the pressure exerted by water on the sides of the container at the bottom is 4606 Pa.
(b)
For the ides of the container halfway down from the top, the height is,
h' = h/2
h' = 0.47/2 = 0.235 m
Then the pressure is obtained as,
[tex]P' = \rho \times g \times h'\\\\P' = 1000 \times 9.8 \times 0.235\\\\P'=2303 \;\rm Pa[/tex]
Thus, the pressure exerted by water on the sides of the container halfway down from the top is of 2303 Pa.
Learn more about the atmospheric pressure here:
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Which formula is used to find an object's acceleration?
Answer:
The answer is C.
Explanation:
A train 4.00 3 102 m long is moving on a straight track with a speed of 82.4 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 16.4 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answer:
The value is [tex]t =29.2 \ s [/tex]
Explanation:
From the question we are told that
Generally the average velocity of the train is mathematically represented as
[tex]v = \frac{u + v}{2}[/tex]
substituting 82.4 km/h for u and 16.4 km/h. for v
[tex]v = \frac{82.4 + 16.4}{2}[/tex]
[tex]v = 49.4 \ km/h[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{ L}{v}[/tex]
substituting 49.4 \ km/h for v and [tex]4.00 * 10^2 \ m = 0.400 \ km[/tex]
[tex]t = \frac{ 0.400}{49.4}[/tex]
[tex]t = 0.00809 \ h [/tex]
converting to seconds
[tex]t = 0.00809 * 3600 [/tex]
[tex]t =29.2 \ s [/tex]
How does the lighter material of the bamboo affect the force created by the rider on the bike?
Newton's second law allows us to find the result for the effect of building a lower mass bike is:
An increase in the acceleration of the bicycle.
Newton's second law states that force is equal to the product of mass and acceleration of the body.
∑ F = m a
Where the bold letters indicate vectors, m is the mass and a the acceleration.
They indicate that the bicycle is made of bamboo, so it has less mass, therefore the acceleration is:
[tex]a = \frac{\sum F}{m}[/tex]
if the mass decreases, more acceleration is recorded, therefore from the kinematic reactions the bicycle reaches higher speeds.
v = v₀ + a t
The bicycles parts of the rest.
v = a t
In conclusion, using Newton's second law we can find the result for the effect of building a bicycle with alower mass is:
An increase in the acceleration of the bicycle.
Learn more here: brainly.com/question/8522449
Which of the following BEST explains the energy transformation that takes place when a battery alarm clock goes off.
A. electrical energy --> chemical energy --> sound energy
B. electrical energy --> chemical energy --> mechanical energy --> sound energy
C. chemical energy --> electrical energy --> mechanical energy --> sound energy
D. it is impossible to tell without knowing what type of battery
I think it might be Either A or B im thinking B
A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 61 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 44 kg?
Answer:
6957.04N
Explanation:
Using
vf2=vi2+2ad
But vf = 0 .
So convert 50km/hr to m/s, and you need to convert 61 cmto m
(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s
61cm * (1m/100cm) = .61m
So n
0 = (13.9m/s)^2 + 2a(.61m)
a = 158.11m/s^2
So
using F = ma
F = 44kg(158.11m/s^2) = 6957.04N
Answer:
6.95 kN
Explanation:
Given that
Speed of the car, u = 50 kmph
Distance moved by the passenger, s = 61 cm = 0.061 m
Mass of the passenger, m = 44 kg
Converting the initial velocity from km/h to m/s, we have
50 kmph = 50 * 1000/3600
50 kmph = 13.89 m/s
Using one of the equations of motion,
v² = u² + 2as, where v = 0,making a the subject of formula
a = -u²/2s
a = -(13.89²) / 2 * 0.61
a = -192.93 / 1.22
a = -158 m/s²
The force acting on the passenger's leg, F = m.a, so
-F = 44 * -158
F = 6952 N, or 6.95 kN
what is the net force on a backpack with a mass of 12.0 kg and an acceleration of 0.5 m/s^2?
Answer:
The answer is
6.0 NExplanation:
The net force acting on an object when given the mass and acceleration can be found by using the formula
Force = mass × accelerationFrom the question
mass of backpack = 12 kg
acceleration = 0.5 m/s²
So the force is
Force = 12 × 0.5
We have the final answer as
6.0 NHope this helps you
A tennis ball is shot vertically upward inside a tower with an initial speed of
20.0 m/s. Neglect air resistance. Approximately how long does it take the tennis ball to reach its maximum height?
0.50 s
2.04 s
4.08 s
6.08 s
9.80 s
Determine the velocity of the ball 3.0 s after it is thrown
9.40 m/s, downward
9.40 m/s, upward
29.4 m/s, downward
38.8 m/s, upward
38.8 m/s, downward
The ball's height at time t is
y = (20.0 m/s) t - 1/2 g t²
where g is the acceleration due to gravity, with magnitude 9.80 m/s².
Also, recall that
v² - u² = 2 a ∆y
where u is the initial velocity, v is the final velocity, a is the acceleration, and ∆y is the change in height. Let Y be the maximum height. At this height, v = 0, so
- (20.0 m/s)² = 2 (-g) Y
==> Y ≈ 20.408 m
Plug this into the first equation and solve for t :
Y = (20.0 m/s) t - 1/2 (9.80 m/s²) t²
==> t ≈ 2.04 s
The ball's velocity at time t is
v = 20.0 m/s - g t
After t = 3.0 s, its velocity will be
v = 20.0 m/s - (9.80 m/s²) (3.0 s)
v = -9.40 m/s
or 9.40 m/s in the downward direction.
What is the difference between an observation and an inference?
Explanation:
observation is something what you could observe from your organs like eyes ears etc and also it is what you observed during an event for an experiment but inference is what you decide to do after observation or an event.
the act of inferring (to derive by reasoning). Observation = an act or instance of noticing or perceiving.
Thank u!
Which is larger, the Sun's pull on Earth or Earth's pull on the Sun?
a. The Sun's pull on Earth is twice as large as Earth's pull on the much larger Sun.
b. They pull on each other equally.
c. The Sun's pull on Earth is larger.
d. Earth's pull on the Sun is larger.
e. There is no pull or force between Earth and the Sun.
At the time when the sun pulls on earth should be large so it should be pulled on each other equally.
Gravitational force:Here the gravitational force that lies between two masses should be proportional with respect to the product of mass that should be divided via the square of the distance that lies between them. The acceleration should be measured when the force should be divided by the mass. Here the two forces should be equal and opposite
Therefore, the option b is correct.
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when the temperature of matter decrease , the particles do what
Answer:
When the temperature decreases the particals start to slow down.
measurement conversions [metric to metric] 568 cm = m
Answer:
5.68 meters
Explanation:
hope this helps!
Answer:
5.68
Explanation:
to convert cm to m you move the decimal point 2x to the left
The volume of water and an egg in a graduated beaker is 200mL. After the egg is removed the volume of the water is found to be 125mL . What is the volume of the egg in cm ^3
Answer:
75 cm³
Explanation:
The volume of the egg is equal to the volume of the egg and water minus the volume of the water.
V = 200 mL − 125 mL
V = 75 mL
V = 75 cm³
Answer:
75 cm
Explanation:
The equation would be 200ml -125 ml thus finding the volume of the egg.
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.A. Which laser has its first maximum closer to the central maximum?B. What is the distance delta ymax-min between the first maxima (on the same side of the central maximum) of the two patterns?C. What is the distance Delta ymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Answer:
Now
Using
ym = m x λ x L/d
therefore for each of first Max we have m =1
And laser 1 had
y = (d/20) x (6.0m)/d
y = 6.0m /20 = 0.3
Then laser 2 will be
y = 6.0m/15 = 0.4
y ( laser1 ) < y ( laser2 )
So the first maxima of laser 1 will be closer to the Central maxima
So
(0.4m -0 .3 m) = 0.1m
( C)
Now for laser 1 we say
y= 26.0 m / 20 = 0.6 m
Laser 2
We have
ym=(m+1/2) x λ x L/d
So
Because there is no central minimum the first minimum is at m = 0
We can way 3rd minimum is at m = 2
So
y = (2.5) x 6.0 / 15 = 1m
So
Δy= 1m - 0.6m = 0.4 m
A falling object satisfies the initial value problem dv dt = 9.8 − v 5 , v(0) = 0 where v is the velocity in meters per second. (a) Find the time that must elapse for the object to reach 95% of its limiting velocity. (Round your answer to two decimal places.) s (b) How far does the object fall in the time found in part (a)? (Round your answer to two decimal places.) m Additional Materials
Answer:
a. t [tex]\simeq[/tex] 14.98 sec
b. x = 501.27 m
Explanation:
From the given information:
[tex]\dfrac{dv}{dt}=9.8-(\dfrac{v}{5 })[/tex] and [tex]v(0)=0[/tex]
[tex]\dfrac{dv}{dt}=\dfrac{49-v}{5 }[/tex]
[tex]\dfrac{dv}{49-v}=\dfrac{dt}{5 }[/tex]
Taking Integral of both sides
[tex]- ln(49-v) = \dfrac{t}{5} + C[/tex]
at t=0 we have v=0
This implies that
[tex]- ln(49-0) = \dfrac{0}{5} + C[/tex]
[tex]C= - ln(49)[/tex]
Thus:
[tex]\dfrac{t}{5} - In (49) = - In (49 -v) \\ \\ In(49) - \dfrac{t}{5} = In (49-v)[/tex]
[tex]49-v = e^{(-\frac{t}{5} +ln(49))}\\ \\ v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]
The limiting velocity when the time is infinite is :
95% of 49 = 46.55
∴
[tex]0.05= e^{(-\dfrac{t}{5})}[/tex]
[tex]\dfrac{t}{5}= In(\dfrac{1}{0.05})[/tex]
[tex]\dfrac{t}{5}=2.9957[/tex]
t = 5 × 2.9957
t [tex]\simeq[/tex] 14.98 sec
b.) [tex]v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]
[tex]v = \dfrac{dx}{dt}=49 - 49e^{(-\dfrac{t}{5})}[/tex]
[tex]dx=(49 - 49e^{(-\frac{t}{5})}) \ dt[/tex]
Taking integral of both sides.
[tex]x = 49t + 245 e^{(\frac{-t}{5})} +C[/tex]
at time t = 0 , distance x traveled = 0
∴
C= - 245
Therefore
[tex]x = 49t + 245 e^{(\frac{-t}{5})} -245[/tex]
replacing the value of t = 14.98
[tex]x = 49(14.98) + 245 e^{(\frac{-14.98}{5})} -245[/tex]
x = 501.27 m
A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340 kPa. The bottle is turned upsidedown and the cap is released so that the water is rapidly forced out of the bottle. If the air in the bottle undergoes an adiabatic pressure change, what is the pressure in the bottle when the bottle is five-sixths full of air
Answer:
The value is [tex] P_G = 2.925 *10^{5} \ Pa[/tex]
Explanation:
From the question we are told that
The volume of the bottle is [tex]v = 2 \ L = 2 * 10^{-3} \ m^3[/tex]
The gauge pressure of the air is [tex]P_g = 340 \ kPa = 340 *10 ^{3} \ Pa[/tex]
Generally the volume of air before the bottle is turned upside down is
[tex]V_a = \frac{3}{4} * V[/tex]
[tex]V_a = \frac{3}{4} * 2 *10^{-3}[/tex]
[tex]V_a = 0.0015 \ m^3 }[/tex]
Generally the volume air when the bottle is turned upside-down is
[tex]V_u = \frac{5}{6} * 2 *10^{-3}[/tex]
[tex]V_u = 0.00167 \ m^3 [/tex]
From the the mathematical relation of adiabatic process we have that
[tex]P_g * V_a^r = P_G * V_u^r[/tex]
Here r is a constant with a value r = 1.4
So
[tex] 340 *10 ^{3} * 0.0015^{1.4} = P_G * 0.00167^{1.4}[/tex]
[tex] P_G = 2.925 *10^{5} \ Pa[/tex]
In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9.0 m above him, releases a ball from rest. While still in flight, the acrobat catches the ball just as it reaches him.Required:If he left the trampoline with a speed of 5.6 m/s, how long is he in the air before he catches the ball?
Answer:
1.6 secs
Explanation:
In a circus act, an acrobat upwards from the surface of a trampoline
At that same moment another acrobat perched 9.0m above him
A ball is released from rest
While still in motion the acrobat catches the ball
He left the ball with a trampoline of 5.6m/s
Since the ball is falling downwards from a distance then acceleration will be negative
a= -9.8
s= d
s= 1/2at^2
= 1/2 × (-9.8)t^2
= 0.5× (-9.8)t^2
d = -4.9t^2
Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows
9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2
9 - 4.9t^2= 5.6t + (-4.9)t^2
9 - 4.9t^2= 5.6t - 4.9t^2
9= 5.6t
t= 9/5.6
t= 1.6 secs
4. What is the instantaneous acceleration at t= 10 s?
Answer:
I am fairly certain the answer is 2m/s^2
Explanation:
An owl is carrying a mouse to the chicks in its nest. It is 4.00 m west and 12.0 m above the center of the 30 cm diameter nest and is flying east at 2.50 m/s at an angle 32° below the horizontal when it accidentally drops the mouse. Will it fall into the nest? Find out by solving for the horizontal position of the mouse (measured from the point of release) when it has fallen the 12.0 m.
Answer:
4.61 m away
Explanation:
No, the owl will not fall into the nest.
The vertical component of the velocity, v of mouse after travelling vertical distance of 12 m is given by
v² - (4 * sin 32)² = 2 * 9.81 * 12
v² - (4 * 0.53)² = 235.44
v² - 4.494 = 235.44
v² = 235.44 + 4.494
v² = 239.934
v = 15.49 m/s
The time taken to travel the said 12 m downwards =
(15.49 - 4 * sin 32) / 9.81
(15.49 - 4 * 0.53) / 9.81
(15.49 - 2.12) / 9.81
13.37 / 9.81 = 1.36 s
Horizontal distance traveled in this time = 4 * (cos 32) * 1.36
= 4 * 0.848 * 1.36
= 4.61 m
Therefore, the horizontal position of the mouse is 4.61 m away from the point of release.
The centroid of the plane region shown is at c. use the method of composite area to determine the radius (R)? please someone help me i nedd your helps guys please!
A group of engineers is preparing a satellite to land by moving it 10% closer to
Earth in each rotation. Which statement is correct about the rotational inertia of
the satellite? (1 point)
O Rotational inertia does not change because it is conserved.
Rotational inertia increases proportional to the decrease in the
radius of rotation.
Rotational inertia first decreases and then increases as the
satellite is ready to land.
O
Rotational inertia decreases proportional to the decrease in the
radius of rotation.
Answer:
Rotational inertia decreases proportional to the decrease in the radius of rotation.
Explanation:
The rotational inertia decreases proportional to the decrease in the
radius of rotation.
Rotational inertia is directly proportional to the angular momentum of the object and inversely proportional to its angular velocity.
[tex]I = \frac{L}{\omega}[/tex]
where;
I is the rotational inertiaL is the angular momentumω is the angular speedThe angular momentum is given as;
[tex]L = mvr[/tex]
where;
m is the mass of the objectv is the velocity of the objectr is the radius of the objectThe new rotational inertia equation becomes;
[tex]I = \frac{mvr}{\omega}[/tex]
From this equation, we can observe that the rotational inertia is directly proportional to the radius of the object.A 10% closer to the Earth, means a decrease in the radius of the satellite by 10%.
Thus, a decrease in the rotational inertia as well.
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Which situation is most likely to lead to improving a scientific theory?
A. Analyzing the history of how a theory was developed
O B. Gathering scientists for a brainstorming session
c. Increasing public awareness of the need for scientific research
O D. Performing experiments that have never been done before
SUBMIT
Performing experiments that have never been done before is most likely to lead to improving a scientific theory. The correct option is D.
What is scientific theory?A theory is a well-thought-out elaboration for natural-world observational data that has been built using the scientific process and incorporates many facts and hypotheses.
A theory not only helps to explain established information; it also allows scientists to predict what they will see if the hypothesis is correct.
Scientific theories can be tested. Emerging data should be consistent with an existing theory.
To improve a scientific theory, conduct novel experiments on it. This will provide greater clarity on the theory.
It may reveal the limitations of this theory, as well as aspects of the theory that can be improved.
Thus, the correct option is D.
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A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −0.0200 m in the same direction as the electric field. The electron's speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.04 ✕ 10−17 J. The electron continues to x = −0.210 m within the constant electric field. If there's a change in potential energy of −9.60 ✕ 10−17 J as it goes from x = 0.190 m to x = −0.210 m, find the electron's speed (in m/s) at x = −0.210 m.
Answer:
The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]
Explanation:
Given that,
Electric field [tex]E=1.50\times10^{3}\ N/C[/tex]
Distance = -0.0200
The electron's speed has fallen by half when it reaches x = 0.190 m.
Potential energy [tex]P.E=5.04\times10^{-17}\ J[/tex]
Change in potential energy [tex]\Delta P.E=-9.60\times10^{-17}\ J[/tex] as it goes x = 0.190 m to x = -0.210 m
We need to calculate the work done by the electric field
Using formula of work done
[tex]W=-eE\Delta x[/tex]
Put the value into the formula
[tex]W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))[/tex]
[tex]W=-5.04\times10^{-17}\ J[/tex]
We need to calculate the initial velocity
Using change in kinetic energy,
[tex]\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2[/tex]
[tex]\Delta K.E=\dfrac{-3mv^2}{8}[/tex]
Now, using work energy theorem
[tex]\Delta K.E=W[/tex]
[tex]\Delta K.E=\Delta U[/tex]
So, [tex]\Delta U=W[/tex]
Put the value in the equation
[tex]\dfrac{-3mv^2}{8}=-5.04\times10^{-17}[/tex]
[tex]v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}[/tex]
Put the value of m
[tex]v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}[/tex]
[tex]v=1.21\times10^{7}\ m/s[/tex]
We need to calculate the change in potential energy
Using given potential energy
[tex]\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})[/tex]
[tex]\Delta U=-4.56\times10^{-17}\ J[/tex]
We need to calculate the speed of electron
Using change in energy
[tex]\Delta U=-W=-\Delta K.E[/tex]
[tex]\Delta K.E=\Delta U[/tex]
[tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}[/tex]
Put the value into the formula
[tex]v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}[/tex]
[tex]v_{f}=1.5\times10^{7}\ m/s[/tex]
Hence, The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]
