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liberstina [14]

2 years ago

8

In attempting to pull a 1500 kg car out of a ditch, Al exerts a force of 200 N for 5 s, Bill exerts a force of 500 N for two s,

and Clyde exerts a force of 300 N for four seconds. The car does not move at all. Who provided the greatest impulse?

1 answer:

MArishka [77]2 years ago

4 0

Answer:

Clyde will provide greater impulse

Explanation:

We have given that Al exerts a force of 200 N for 5 sec

We know that impulse is given by, impulse = force ×time = 200×5 =1000 N-s

Bill exerts a force of 500 N for 2 sec

So impulse = 500×2 = 1000 N-s

Now the force exerted by Clyde 300 N for 4 sec

So impulse = 300×4 = 1200 N-s

From above calculation we can see that Clyde provide greater impulse than any other

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At a given time of day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4-f

katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

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shadow length =2.8 ft

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let the height of tree be h

as $\theta$ will remain same thus

$tan\theta =\frac{h}{24.64}$

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8 0

3 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago

Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point

Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

$B = \frac{\mu_oI}{2\pi R}$

At the center, the magnetic field strength is twice

$B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}$

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

$B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A$

Therefore, current needed is 39.8 A

6 0

2 years ago

A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object

Darya [45]

Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, $F_x=8\ N$

Force acting along +y direction, $F_y=6\ N$

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

$R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}$

Here $\theta=90^{\circ}$

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{8^2+6^2}$

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.

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3 years ago

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Hatshy [7]

What are the options?

4 0

2 years ago