All categories

3 years ago

What is 1 and 2 I need it asap

Physics

1 answer:

ExtremeBDS [4]3 years ago

5 0

The correct answer for the first one is A) Breaking rocks into smaller pieces. The correct answer to the second one is A) It has more energy. Hope this helps.

You might be interested in

Please help me plzzzzz

Andrews [41]

Answer:

what

Explanation:

6 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

4 years ago

Katie works in a textile industry. In what sector does Katie work?

soldi70 [24.7K]

Textiles are what we buy to make clothing. It could be the Fashion sector or Production , Trade and or Manufacturing goods.

6 0

3 years ago

To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle

Alenkinab [10]

Answer:

Hello your question is incomplete attached below is the missing part of the question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I

answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L

part A = attached below

Explanation:

Part A :

Assuming that mass of swing is negligible

α = T/I

where ; T = torque, I = inertia,

hence T = L/2*9*(M1 - M2)

also; I = $M1*(L/2)^2 + M2*(L/2)^2$ = ( M1 + M2) * (L/2)^2

Finally the magnitude of the angular acceleration α

α = 2*[(M1 - M2)/(M1 + M2)]*g/L

Part B attached below

7 0

3 years ago

An atom of hydrogen consists of one _______________ surrounded by one _________________. A) proton, neutron B) neutron, proton C

maksim [4K]

This would be C), one proton surrounded by one electron :)

5 0

3 years ago

Read 2 more answers