Answer:
T²= 4π²R³/GM
Explanation:
First we know that
Fg= Fc
Because centripetal force must equal gravitational force
So
GMm/R² = Mv²/R
But velocity is 2πR/T
So by substitution we have
GMm/R²= M (2πR/T)/T
We have
T²= 4π²R³/GM as period
Starting from rest, a car accelerates at a rate of 7.8m/s^2 for 4.9 seconds. What is it’s velocity at the end of this time?
Answer:
7.8 m/'s = Change of accelerates / time taken =7.8/4.9=
Explanation:
please solve that answer
An open freight car rolls friction-free along a horizontal track in a pouring rain that falls vertically. As water accumalates in the car, its speed
Answer:
decrease
Explanation:
weight
As water accumulates in the car, its speed decrease according to Newton's second law of motion
What is newton's second law of motion?Newton's second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
according to newton's second law
force = mass * acceleration
as water accumulate in the car , mass will increase
since mass and acceleration are inversely proportional to each other
hence , acceleration will decrease and speed will decrease
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An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of the transverse and longitudinal waves in rock are 8.8 km/s and 5.9 km/s respectively. A seismograph records the arrival of the transverse waves 69 s before that of the longitudinal waves. Assuming the waves travel in straight lines, how far away is the center of the earthquake
Answer:
1239.216 km
Explanation:
The speed of the transverse = 8.8km/s
The speed of the longitudinal = 5.9km/s
distance = speed x time,
8.8km/s x trans_time = 5.9km/s x long_time
8.8 / 5.9 = long_time / trans_time
1.49 = long_time / trans_time
long_time = 1.49 trans_time
the transverse wave was 69s faster than longitudinal,
trans_time - long_time = 69s
trans_time - 1.49trans_time = 69s
0.49 trans_time = 69
trans_time = 69 / 0.49 = 140.82s
long_time = 140.82 - 69 = 71.82s
the distance of the earthquake;
distance = 8.8 x 140.82 = 1239.216 km
1.
Which of the follow
following
cannot be zero?
A. Distance B. Displacement
" Velocity D. Acceleration
C.
what would the answer be ?
Cell phones require powerful batteries in order to work effectively. Which
activity is best described as an engineering endeavor related to cell-phone
batteries?
A. Designing an investigation that will help determine where the
materials used to make them can be found
B. Performing experiments to explain how different materials used in
batteries store energy
O C. Developing a plan to convince cell-phone companies to use their
batteries
D. Improving their design to make it easier to recycle them when they
are no longer useful
The activity D. Improving their design to make it easier to recycle them when they are best described as an engineering endeavor related to cell phone batteries.
What is an engineering endeavor related to the generation of a product?An engineering endeavor related to the generation of a product can be defined as any type of solution in order to enhance a particular feature of a given product, thereby increasing its utility in order to maximize the process that leads to its use by the clients of such product and or service in the market.
Endeavor engineering generally is based on the combination between extensive discovery and product development in order to solve a particular issue related to the product.
Therefore, with this data, we can see that an engineering endeavor related to the generation of a product is aimed at enhancing a particular feature of such a product.
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What are the benefits of living in a country with a growing population rate?
Answer:
The benefits are the place you live in will have more oppurtunities for new jobs as well as growth in residential areas.
Explanation:
A bus is moving at a speed of 20 m/s, when it begins to slow at a constant rate of 5 m/s2 in order to stop at a bus stop. If it comes to rest at the bus stop, how far away was the bus from the stop?
Answer:
u=20m/sec
v=0
a=5m/sec^2
v^2=u^2-2as[ here acceleration is negative]
0=400-2x5xs
-400= -10s
s= 40 metre.
The bus stop is 40 m away from the point at which it starts to slow down at -5 m/s².
State third equation of motion?
The third equation of motion is -
v² - u² = 2aS
Given is a bus moving at a speed of 20 m/s. It begins to slow at a constant rate of 5 m/s² in order to stop at a bus stop.
We can write -
[u] = 20 m/s
[a] = - 5 m/s²
[v] = 0 m/s
Using the third equation of motion -
v² - u² = 2aS
- (20)² = 2 x - 5 x S
- 400 = - 10 S
S = 400/10
S = 40 m
Therefore, the bus stop is 40 m away from the point at which it starts to slow down at -5 m/s²
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For an object that experiences constant non-zero acceleration, V f = V i
True or False
Answer:false
Explanation:
Which is not standing in the way of astronomers getting a good view of distant stars? A. some stars are too far away for our telescope to see B. lights get distorted by gas and dust as it goes through the atmosphere C. Redshift makes stars difficult to see D. there is alot of light pollution on earth
Answer: D. There is a lot of light pollution on earth
Explanation: The light pollution on Earth has nothing to do with the stars in the sky
Answer: C. Redshift makes stars difficult to see.
Explanation:
I did the test
A woman on a snowmobile moving with a constant velocity east down the road fires a flare straight upward and the snowmobile continues to move with a constant
velocity as the flare is in the air. Assume no air resistance. Where will the flare land?
A) it will hit the person on the snowmobile who fired the flare
B) Somewhere behind the snowmobile, depends on velocity
C) in front of the snowmobile
D) Impossible to know
E) behind the snowmobile in exactly the same location (relative to the ground) from which it was fired
A machine carries a 100kg cargo to a boat at a rate of 10m/s2. The distance between the ground to the boat is 50ft. If the machine must transfer the cargo to the boat in 5 minutes, how much power should the machine exert?
Answer:
50.8 watt
Explanation:
we know that P=W÷t
W=F.S S-->distance=50 ft= 15.24 m
F=ma
=100×10=1000 N
SO W= 1000×15.24
=15240 J
NOW
P=W÷t t=5 mints = 5×60=300 sec
P=15240÷300
P=50.8 watt
If a moving clock is ticking half as fast as normal what speed is the clock traveling?
Answer:
Speed of moving clock is
[tex]V2[/tex] =πr/(86400) m/s
Here,
m and s are SI units of distance and time respectively.
Explanation:
If
radius of circular clock= r
than,
total distance covered on clock=S=2πr m
here, m=SI unit of distance
and required time for covering the total distance=t= 86400s
speed of normal clock=[tex]V1[/tex]=S/t
[tex]V1[/tex]=2πr/86400 m/s
As, moving moving clock is ticking half as fast as normal clock so,
speed of moving clock=[tex]V2[/tex]=[tex]V1[/tex]/2
[tex]V2[/tex]=2πr/(86400)*2 m/s
[tex]V2[/tex] =πr/(86400) m/s
In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min , 43.6 s43.6 s . Suppose "Pre" was at the 7.85 km7.85 km mark at a time of 25.0 min25.0 min . If he accelerated for 60 s60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s60 s interval. Assume his instantaneous speed at the 7.85 km7.85 km mark was the same as his overall average speed up to that time.
Answer:
a = 0.161 [tex]$m/s^2$[/tex]
Explanation:
Given :
[tex]$ d_{total}[/tex] = 10 km = 10000 m
[tex]$t_{total} $[/tex] = 27 min 43.6 s
= 1663.6 s
[tex]$d_1$[/tex] = 7.85 km = 7850 m
[tex]$t_1$[/tex] = 25 min = 1500 s
[tex]$t_2$[/tex] = 60 s
Now the initial speed for the distance of 7.85 km is
[tex]$ v_1 = \frac{d_1}{t_1} = \frac{7850}{1500}$[/tex] = 5.23 m/s
The velocity after 60 s after the distance of 7.85 kn is
[tex]$v_2 = v_1 + at_2$[/tex]
= 5.23 + a(60)
The distance traveled for 60 s after the distance of 7.85 km is
[tex]$d_2 = v_1t_2+\frac{1}{2}at_2^2$[/tex]
[tex]$d_2 = (5.23)(60)+\frac{1}{2}a(60)^2$[/tex]
= 313.8 + a(1800)
The time taken for the last journey where the speed is again uniform is
[tex]$t_3 = t_{total}-t_1-t_2 $[/tex]
= 1663.6 - 1500 - 60
= 103.6 s
Therefore, the distance traveled for the time [tex]$t_3$[/tex] is
[tex]$ d_3 = v_2 t_3$[/tex]
= (5.23+60a)(103.6)
= 541.8 + 6216 a
The total distance traveled,
[tex]$ d_{total}= d_1 + d_2 + d_3$[/tex]
Now substituting the values in the above equation for the acceleration a is
10000 = 7850 + (313.6 + 1800a) + (541.8 + 6216a)
10000 = 8706.5 + 8016a
1294.4 = 8016a
a = 0.161 [tex]$m/s^2$[/tex]
What is the force applied by the ground called? A. applied force B. friction C. gravity
Answer:
gravity i think hope this helps
Explanation:
Is it possible to accelerate and not speed up or slow down?
Answer: No,
explanation: When the object is neither speeding up or slowing down, we can say that its speed is constant.
Hope this helps
Plz mark brainlesit
A student is sitting at rest in a chair. How does the force that the student exerts
on the chair compare to the force the chair exerts on the student? *
•a larger magnitude but the opposite direction
•a smaller magnitude but the same direction
•the same magnitude but the opposite direction
•the same magnitude and the same direction
Answer:
the same magnitude but the opposite direction
Explanation:
Newton's third law of motion states that there is always an equal and opposite reaction to every action. This means that the amount of force exerted upon an object is equal to the amount of force the object exerts but in an opposite direction.
This is the case in this scenario where a student sits at rest in a chair. The student is supplying the action force being exerted on the chair. According to the third law of Newton, the chair will exert the same size of force back in the student but in an opposite direction.
Hence, the force the chair exerts on the students compare with that of the student in the sense that they are the same magnitude (size) but the opposite directions.
The slope of the x-t curve at any point represents:_______a. Displacementb. Velocityc. Accerlationd. None of these
Answer:
Velocity
Explanation:
x-t curve means position vs time graph in which position (x) is in x-axis and time is in y-axis.
Slope of a graph = [tex]\dfrac{\Delta y}{\Delta t}[/tex].
We know that, velocity = displacement/time
or we can say that slope = displacement/time = velocity
Hence, the correct option is (b) "velocity".
You are walking at 3.75 km per hour across a frozen lake in the snow. You do not realize that with each step you turn 0.350 degrees to your right. If your step length is 74.0 cm what is the diameter, in meters, of the circle that you are inadvertently tracing out?
Answer:
242.27929622673 meters
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]
which of the following has the greatest inertia ping pong ball, golf ball, softball, and a bowling ball
Answer:
bowling ball
Explanation:
A bowling ball has more mass than the others, thus having more inertia.
HELLO CAN SOMEONE HELP ME PLS
A car is moving at 35 mph and comes to a stop in 5 seconds.
Find the acceleration of the car.
Answer:
I do believe it's 7
when the temperature of matter decrease , the particles do what
Answer:
When the temperature decreases the particals start to slow down.
Tammy jogged 350 meters at a velocity of 7 m/s. How long did it take her to jog this distance?
Answer:
50 seconds
Explanation:
divide 350 by 7...
add my tik(tok) slapddaddy
A tennis ball is hit into the air and moves along an arc. (a) Neglecting air resistance, where along the arc is the speed of the ball a minimum? (Select all that apply.) at the initial position of motion at the highest point in the trajectory at the final position of motion
Answer:
at the highest point in the trajectory
Explanation:
When the tennis ball is hit, it moves in the air along a curve or an arc. This path is the parabola curve. Such a motion in the two dimension is known as projectile motion. It is constant accelerated motion in the downward direction.
The velocity of the ball is minimum at the highest point of the motion. When we hit the ball, the ball moves up to certain eight and then it gradually fall back to the earth surface along a curve.
The horizontal velocity of the ball is always the same along the curve. Only the vertical velocity varies. As the ball reaches the top of the curve or the maximum height, its vertical velocity becomes zero.
So, speed of the tennis ball is minimum at the highest point of the path.
A ball is thrown straight upward with a speed of 36 m/s. How long does it take to return to its starting point, assuming negligible air resistance?
Answer:
The time taken for the ball to return to the starting point is is 7.4 s
Explanation:
Given;
initial velocity of the ball, u = 36 m/s
the final vellocity at maximum height, v = 0
let time taken for the ball to reach maxmimum height = t
Time taken for the ball to return to the starting point is known as time of flight, calculated as;
[tex]t = \frac{v-u}{-g} \\\\T = 2t\\\\T = \frac{2(0-u)}{-g}\\\\T = \frac{-2u}{-g}\\\\ T = \frac{2u}{g}[/tex]
T = (2 x 36) / 9.8
T = 7.4 s
Therefore, the time taken for the ball to return to the starting point is is 7.4 s
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 mass on a string of unknown length oscillates as a pendulum with a period of 1.8 s. What is the period in the following situations? (Parts (a) to (d) are independent questions, each referring to the initial situation.)(a) The mass is doubled?(b) The string length is doubled?(c) The string length is halved?(d) The amplitude is doubled?
Answer:
Using
Period ( P) is given as
P~√(L/g).
a) since mass has no effect on the period of a pendulum. So, the period will remain 1.8seconds
b) using the formula above ,period varies with the square root of the length. Thus , when the length doubles, the period is multiplied by √2. So, the period is 1.8s*√2 = 2.54s
c) in this case, the period is multiplied by √(1/2).
1.8√(1/2)=1.27s.
d) amplitude of the pendulum doesn't affect the period (unless itsvery high, so, the period is still 1.8s
How does gamma radiation differ from alpha or beta particle radiation?
1) it does not consist of matter
2) it only consists of space
3) it does not consist of energy
4) it only consists of matter
Answer:
1.
Explanation:gamma rays are the most powerful in the electromagnetic spectrum and they are a result of a radioactive atom.they aren't made of matter but just energy as a wave.
PLEASEEEEEE HELP
A jet makes a landing traveling due east with a speed of 120 m/s. If the jet comes to rest in 13.5 s , what’s the magnitude of its average acceleration?
Part B: What is the direction of its average acceleration? (North, south, east, west??)
Answer:
8.89 m/s² west
Explanation:
Assume east is +x. Given:
v₀ = 120 m/s
v = 0 m/s
t = 13.5 s
Find: a
v = at + v₀
0 m/s = a (13.5 s) + 120 m/s
a = -8.89 m/s²
a = 8.89 m/s² west
