All categories

3 years ago

You are accelerating upwards in an elevator when the net force on you increases. What happens to the acceleration

Physics

1 answer:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

the acceleration of the elevator is increasing

Explanation:

For this exercise we propose the solution using Newton's second law

F -W = m a

F = m (g + a)

If the net force increases, it implies that the acceleration of the elevator is increasing, since the acceleration of gravity is constant as the ascent is accelerating.

You might be interested in

if a certain car, going with speed v1, rounds a level curve with a radius r1, it is just on the verge of skidding. if its speed

schepotkina [342]

Answer:

The correct answer is 4R1

Explanation:

According to the given scenario ,the radius of the tightest curve on the same road without skidding is as follows:

As we know that

Centeripetal Acceleration is

= v^2 ÷ r

In the case when velocity becomes 2 times so the r would be 4 times

So, the radius of the tightest curve on the same road without skidding is 4R1

3 0

4 years ago

A ball whose mass is 0.4 kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 6 m/s. Collapse question pa

Colt1911 [192]

Answer:

The the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

Explanation:

Given:

Mass of ball $m = 0.4$ kg

Initial speed $v_{i} = -8 \frac{m}{s}$

Rebound speed $v_{f} = 6 \frac{m}{s}$

Contact time interval $\Delta t = 0.5 \times 10^{-3}$ sec

For finding the average magnitude of the force on the ball by the floor is given by,

$F_{avg} = \frac{\Delta P}{\Delta t}$

Here $\Delta P = m (v_{f}- v_{i} )$

$F_{avg} = \frac{m (v_{f} -v_{i} )}{\Delta t}$

$F_{avg} = \frac{0.4 \times (6 -( -8 ) )}{0.5 \times 10^{-3} }$

$F_{avg} = 11.2 \times 10^{3}$ N

Therefore, the the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

6 0

3 years ago

Read 2 more answers

A small asteroid with a mass of 1500 kg moves near the earth. At a particular instant the asteroid’s velocity is ⟨3.5 × 104, −1.

zalisa [80]

Answer:

$P_{f}$ =(5.7 x $10^{7$ i - 2.24 x $10^{7$ j) kgm/s

Explanation:

Due to earths gravity, force on asteroid is given by:

$F= \frac{Gm_{1}m_{2} }{r^{2} }$ r^

Plugging in the values, we have

F= [(6.67x $10^{-11}$ )(1500)(5.97 x $10^{24}$ )(8x $10^{6}$ i + 9x $10^{6$ j)] / ((8x $10^{6}$ )² + (9x $10^{6$ )² $)^{1.5}$

F= 2736 i^ + 3078 j^

In order find the final momentum of the Asteroid, apply impulse momentum theorem

$P_{f}$ = $P_{i$ + FΔt

$P_{f}$ = 1500(3.5 x $10^{4$ i - 1.8x $10^{4$ j) + (2736i + 3078j)(1.5x $10^{3$ )

$P_{f}$ =(5.7 x $10^{7$ i- 2.24 x $10^{7$ j)kgm/s

4 0

4 years ago

The color of a star is

never [62]

D determined by its temperature

8 0

4 years ago

Read 2 more answers

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago