Need Help with this physics question, just a written response

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

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

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;

The angular momentum is given as;

[tex]L = mvr[/tex]

where;

The new rotational inertia equation becomes;

[tex]I = \frac{mvr}{\omega}[/tex]

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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Answer:

gravity i think hope this helps

Explanation:

Answer:

decrease

Explanation:

weight

As water accumulates in the car, its speed decrease according to 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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#SPJ3

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.

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.

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]

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]

Answer:

I do believe it's 7

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

Answer:

The answer is C.

Explanation:

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]

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".

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

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.

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Answer:

The answer is

Explanation:

The net force acting on an object when given the mass and acceleration can be found by using the formula

From 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

Hope this helps you

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]

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

Answer:

Hope it helps

A Brainliest please

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

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.

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:

Answer:

I am fairly certain the answer is 2m/s^2

Explanation:

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

Answer:

When the temperature decreases the particals start to slow down.

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

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.)