The chances of being involved in an auto collision once a year are

The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s

just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c) 78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0

3 years ago

How is Pluto currently classifed?

Zanzabum

Answer:

Right now Pluto is classified as a <em>dwarf planet.</em>

Explanation:

4 0

3 years ago

Read 2 more answers

A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it

anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

= 0.779

So that;

the average coefficient of static friction = $\frac{sum of coefficient of static friction}{number of trials}$

= $\frac{0.779}{6}$

= 0.12983

The average coefficient of static friction is 0.130

8 0

3 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and

puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

$m_1=43.7 kg$

$m_2=12.1 kg$

$g=9.8 m/s^2$

$a=\frac{1}{2}(9.8)=4.9 m/s^2$

We have to find the mass of the pulley.

According to question

$T_2-m_2 g=m_2 a$

$T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N$

$T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N$

Moment of inertia of pulley= $I=\frac{1}{2}Mr^2$

$(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)$

Where $\alpha=\frac{a}{r}$

$(177.87-214.13)=-\frac{1}{2}(4.9)M$

$-36.26=-\frac{1}{2}(4.9)M$

$M=\frac{36.26\times 2}{4.9}=14.8 kg$

Hence, the mass of the pulley=14.8 kg

6 0

3 years ago

What is the energy of an electromagnetic wave that has a frequency of 5.0 x 10⁵ Hz?

Valentin [98]

Answer: 3.3 x 10^-28 J

Explanation: just answered on a quiz!! :)

5 0

3 years ago