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2 years ago

A car accelerates from a stand still to 75 m/s in 15 seconds what is the acceleration

Physics

2 answers:

inysia [295]2 years ago

8 0

Answer: a = 75 m/s / second

Explanation:

Sphinxa [80]2 years ago

7 0

Answer:

5 m/s^2

Explanation:

acceleration = change in velocity / change in time = Δv / Δt

= 75 m/s / 15 s = 5 m/s^2

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Please i dont understand

valentina_108 [34]

Give me points I will help you

3 0

3 years ago

Calculate the momentum of a 1950 kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant's momentum with

prisoha [69]

Answer:

a) (momentum of the elephant), p1 = 14625 kg×m/s

b) p1 = 609.4×p2

c) (momentum of the hunter), p3 = 720 kg×m/s.

Explanation:

a) if m = 1950 kg is the mass of the elephant and V = 7.50 m/s be the speed of the elephant, then the momentum is given by:

p1 = m×V = (1950)×(7.50) = 14625 kg×m/s

b) if m = 0.0400 kg is the mass of the bullet and v = 600 m/s be the speed of the bullet, then the momentum of the bullet is given by:

p2 = m×v = (0.0400)×(600) = 24 kg.m/s

then:

p1/p2 = (14625)/(24) = 609.4

that is, p1 = 609.4×p2

Therefore, the momentum of the elephant is 609.4 times the momentum of the bullet.

c) the momentum og the hunter after missing the elephant is given by:

p3 = m×v = (90.0)×(8) = 720 kg.m/s

Therefore, the momentum of the hunter is 720 kg×m/s.

6 0

4 years ago

I need help on this question.

stira [4]

Answer:

<em>Answer: B. 49 N</em>

Explanation:

<u>Friction Force </u>

When an object is at rest on a rough surface, it encounters a friction force that opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

$Fr=\mu_k N$

Where $\mu_k$ is the coefficient of static friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

Thus, the friction force is:

$Fr=\mu_k W$

We are given the coefficient of static friction $\mu_k=0.5$ and the mass of the object m=10 Kg. The minimum force needed to make it start moving is the maximum friction force:

Fr=0.5*10*9.8

Fr = 49 N

Answer: B. 49 N

8 0

3 years ago

A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

The displacement of the particle is given by

$d=2.2i+cj$

While the force applied on the particle is

$F=3.2i-3.8 j$

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

$W=F\cdot d$ (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

$0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85$

In this problem, the work done on the particle is

$W=4.0 J$

The force and displacement are still

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

$W=F\cdot d$

$4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8$

In this problem instead, the work done on the particle is negative:

$W=-1.8 J$

As before, the force and displacement are

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

And so again, we calculate the scalar product between force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

$W=F\cdot d$

$-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33$

7 0

3 years ago

Do action and reaction cancel each other out?

galben [10]

Answer:

Action and reaction forces don't cancel each other out because they act on separate objects. ... Action and reaction forces are always equal in magnitude, so it's not possible to exert more force on an object than it can exert back.

Explanation:

8 0

3 years ago