All categories

3 years ago

What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s

Physics

1 answer:

Nikolay [14]3 years ago

8 0

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

You might be interested in

An HR manager is asked to meet with upper management to discuss the

amm1812

Answer:

the answer to this question perhaps is service

3 0

3 years ago

Read 2 more answers

What does it mean for a reaction to be endothermic? choose 2 of the following

Rainbow [258]

Answer: I would choose options C and D

Explanation:

3 0

2 years ago

Read 2 more answers

For a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass 1000 kg. the truck then

babunello [35]

ANSWER

Both trucks will move together with speed v = 6.67 m/s

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

EXPLANATION

As we know that there is no external force on the system of two trucks

So here momentum of the two trucks before collision and after collision will remain same

So here we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

so here we will have

$v_{1i} = 10 m/s$

$v_{2i} = 0$

$m_1 = 2000 kg$

$m_2 = 1000 kg$

now we will have

$2000\times 10 + 1000\times 0 = (2000 + 1000) v$

$v = \frac{20000}{3000} = 6.67 m/s$

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

3 0

2 years ago

Read 2 more answers

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

2 years ago

Which two statements correctly describe the role of a semiconductor in a

KIM [24]

Answer:

It's A and C

7 0

2 years ago