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2 years ago

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Question 1 of 10

1 answer:

kirill115 [55]2 years ago

4 0

Answer:

B) trends method

I'm very sure of this answer

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Carter pushes a bag full of basketball jerseys across the gym floor. The he pushes with a constant force of 21 newtons. If he pu

Zolol [24]

Carter needs a power equal to 63 W to be able to push the bag full of Jersey. This is by using the formula: Power is equal to the product of Force applied and Displacement all over time traveled.

8 0

3 years ago

Read 2 more answers

1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?

Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

$Acceleration = \frac{Change in velocity}{Time taken}$

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

3 0

3 years ago

Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 4R. System A consists of two of

Harman [31]

Answer:

4 smaller disks

Explanation:

We are given;

Mass of smaller and larger disks = M

Radius of smaller disk = R

Radius of larger disk = 4R

Formula for moment of inertia about cylinder axis is:

I = ½MR²

Thus;

For small disk, I_small = ½MR²

For large disk, I_large = ½M(2R)² = 2MR²

We are told that moment of inertia of System A consists of two of the larger disks. Thus;

I_A = 2 × I_large = 2 × 2MR²

I_A = 4MR²

We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;

I_B = I_large + n(I_small)

Where n is the number of smaller disks.

I_B = 2MR² + n(½MR²)

I_B = MR²(2 + n/2)

We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;

I_A = I_B

So;

4MR² = MR²(2 + n/2)

MR² will cancel out to give;

4 = 2 + n/2

Multiply through by 2 to give;

8 = 4 + n

n = 8 - 4

n = 4

5 0

3 years ago

If two forces are in the same direction then do they cancel each other out

stellarik [79]

Im pretty sure that they do

3 0

3 years ago

A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe

tekilochka [14]

Answer:

Angular momentum, $L=6.47\times 10^{-3}\ m$

Explanation:

It is given that,

Radius of the axle, $r=3.21\ mm=3.21\times 10^{-3}\ m$

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

$\tau=\dfrac{dL}{dt}$

Torque acting on the top is given by :

$\tau=F\times r$

Here, F is the tension acting on it. Torque acting on the top is given by :

$\tau=2F\times r$

$2T\times r=\dfrac{L}{t}$

$L=2T\times r \times t$

$L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32$

$L=6.47\times 10^{-3}\ m$

So, the angular momentum acquired by the top is $6.47\times 10^{-3}\ m$ . Hence, this is the required solution.

7 0

3 years ago