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Kazeer [188]
3 years ago
7

An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

onstant acceleration, it travels a distance x. How far would the automobile travel in coming to rest if it had the same acceleration but an initial speed 2v ?
Physics
1 answer:
Molodets [167]3 years ago
3 0

Answer:

4x

Explanation:

Use v^{2} = u^{2} +2as to do the question.

For first instance,

0 = v^{2} +2ax -------------------( 1 )

for second instance,

0 = (2v)^{2} +2as-----------------( 2 )

So by (1) and (2),

s = 4x

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A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0
3 years ago
what causes places near the equator to receive more direct sunlight than the poles? the poles are farther away from the sun than
Andru [333]
This is due to the tilt of the earth on its axis. Although the Sun shines on Earth, because of how the Earth is tilted, the equator is more directly hit compared to places found on the poles. The poles are hit at an angle, therefore the sunlight they receive is lesser than the places at the equator.
5 0
3 years ago
After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0
3 years ago
PLEASE ANSWER
Irina-Kira [14]

Answer:

Farm = 98.1 [N]

Explanation:

To solve this problem we must draw the respective free body diagram, with the forces acting on the monkey. An analysis of the sums on the y-axis must be performed, in this axis the weight is acting down and the forces of both arms pulling up.

Weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 20 [kg]

g = gravity acceleration = 9.81 [m/s²]

W = 196.2 [N] (units of Newtons)

As this force points down, the force of both arms must go up, therefore each arm exerts a force of:

Farm = 196.2 / 2

Farm = 98.1 [N]

8 0
3 years ago
At what speed must the electron revolve round the nucleus of
EleoNora [17]

Explanation:

I think this is it, give it a try

6 0
3 years ago
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