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3 years ago

What is the acceleration of the motorcycle in section CD?

Physics

1 answer:

jenyasd209 [6]3 years ago

6 0

Answer:

30m/s²

Explanation:

The acceleration of the motorcycle in section CD is the slope of the velocity - time graph.

Given that;

Initial velocity = 20m/s Time = 6s

Final velocity = 50m/s Time = 7s

The time interval = 7s - 6s = 1s

So;

Acceleration = $\frac{Change in velocity }{time}$ = $\frac{50 - 20}{1}$ = 30m/s²

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Help pls ASAP thank you

tino4ka555 [31]

Answer: 3rd option

Explanation: Since it has less mass and same force as other object it will have higher acceleration

7 0

3 years ago

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Two train whistles, and , each have a frequency of 392 Hz. is stationary and is moving toward the right (away from ) at a speed

trasher [3.6K]

Answer:

Part a)

f = 371.1 Hz

Part b)

f = 417.7 Hz

Part c)

beat frequency = 46.6 Hz

Explanation:

Part a)

Due to doppler's Effect the frequency of the sound heard by the train which is moving away from the observer is given as

$f_1 = f_0\frac{v + v_o}{v + v_s}$

$f_1 = 392(\frac{340 + 15}{340 + 35})$

$f_1 = 371.1 Hz$

Part b)

Now from the second train which is approaching the person we can say

$f_2 = f_0\frac{v - v_o}{v - v_s}$

$f_2 = 392(\frac{340 - 15}{340 - 35})$

$f_2 = 417.7 Hz$

Part c)

As we know that beat frequency is the difference in the frequency from two sources

$f_b = f_2 - f_1$

$f_b = 417.7 - 371.1 = 46.6 Hz$

8 0

3 years ago

A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea

Kitty [74]

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L² ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223 (-9)

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

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3 years ago

A particle in uniform circular motion requires a net force acting in what direction? Group of answer choices In the direction of

Annette [7]

Answer:

Towards the center of the circle

Explanation:

When an object is moving in uniform circular motion, it means that it is moving at a constant speed in a circular path.

In order to follows a circular path, the particle must have a net force acting on it, that it makes it constantly changing direction - and therefore, causing an acceleration - and this force is called centripetal force.

The direction of this centripetal force is always towards the centre of the circular path.

The nature of the centripetal force is always different, depending on the situation. For instance: for a satellite orbiting the Earth in circular motion, the centripetal force is provided by the force of gravity. For a car moving round a curve, the centripetal force is provided by the force of friction between the tyres and the road.

The magnitude of the centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the object

v is its speed

r is the radius of the circle

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3 years ago

Error found in a variable inertia bar experiment bar

Reil [10]

Answer: ok so u have to add the variable interia by the experiment bar to get ur total

Explanation:

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3 years ago

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