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Evgesh-ka [11]
3 years ago
6

The speed of the cat

Physics
1 answer:
klemol [59]3 years ago
3 0

You awser is c to the question

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3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What
WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

F_{net}=ma

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction F_f, opposing the motion of the box, so to the left

So we can write the net force as

F_{net}=F-F_f

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

a=0

Therefore

F_{net}=0

WHich means:

F-F_f=0

And therefore,

F_f=F=25 N

which means that the force of friction is also 25 N.

6 0
3 years ago
Help me, please
Mars2501 [29]
Wish I could help you you rock
6 0
2 years ago
Amadeo the Archer is practicing his archery skills. He shoots both arrows into a field. His first arrow (arrow A) is embedded ha
attashe74 [19]

Answer:

1

The arrow with greater impart is Arrow B

2

The both arrows will feel the same impulse

Explanation:

1. Arrow B since KE= W = F * d  it used more force to stop itself in a shorter distance.

2. They should feel the same impulse since the both had the same momentum    m * v= F * t

4 0
2 years ago
Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
Drupady [299]

Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

F_K=u_k*m*g

u_K*m*g=m*a_c

u_K*g=a

u_K=\frac{a_c}{g}

a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}

u_K=0.862

4 0
3 years ago
What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? How far has the car traveled when it
nalin [4]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 60 mph = 26.8 m/s

time t = 10 s

Let a be the acceleration and s be he distance traveled.

Use first equation of motion

v = u + a t

26.8 = 0 +  a x 10

a = 2.68 m/s

Use second equation of motion

s = ut + 1/2 at²

s = 0 + 0.5 x 2.68 x 10 x 10

s = 134 m

As, 1 m = 3.28 ft

So, s = 134 x 3.28 ft

s = 439.6 ft

7 0
3 years ago
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