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GrogVix [38]
3 years ago
7

Two men decide to use their cars to pull a truck stuck in the mud. They attach ropes and one pulls with a force of 615 N at an a

ngle of 31◦with respect to the direction in which the truck is headed, while the other car pulls with a force of 961 N at an angle of 25◦with respect to the same direction. What is the net forward force exerted on the truck in the direction it is headed? Answer in units of N.
Physics
1 answer:
stiv31 [10]3 years ago
4 0

Answer:

1398.12 N

Explanation:

We define the x-axis in the direction parallel to the movement of the truck  on and the y-axis in the direction perpendicular to it.

x-components  of the ropes forces

T₁x = 615N*cos31°=527.1579 N  :Tension in direction x of the rope of the car 1

T₂x= 961 N*cos25°=870.96 N  :Tension in direction x of the rope of the car 2

Net forward force exerted on the truck in the direction it is headed (Fnx)

Fnx = T₁x  + T₂x

Fnx = 527.1579 N  + 870.96 N

Fnx = 1398.12 N

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An electric motor moves a chain that pulls the roller coaster car 180 meters to the top of the first hill. The chain exerts a fo
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Xavier is roller skating at 14 km/h and tosses a set of keys forward on the ground at 8 km/h. The speed of the keys relative to
Arturiano [62]

Answer:

22 km/h

Explanation:

Given that,

Speed of Xavier, v = 14 km/h

He tosses a set of keys forward on the ground at 8 km/h, v' = 8 km/h

We need to find the speed of the keys relative to the ground. Let it is V.

As both Xavier and the keys are moving in same diretion. The relative speed wrt ground is given by :

V = v+v'

V= 14 + 8

V = 22 km/h

So, the speed of the keys relative to the ground is 22 km/h.

4 0
2 years ago
8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of ball.

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of bottle.

We are given:

m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s

Putting values in above equation, we get:

(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0
3 years ago
Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.
Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0
3 years ago
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