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GaryK [48]
3 years ago
15

Jerry is conducting an experiment to test friction. His setup is shown below. If he wants to test the friction between different

types of materials, what should he change in the setup above? A. Replace the mass with a different size mass B. Replace the string with a different length string C. Replace the spring scale with a different kind of scale D. Replace the wooden block with a different kind of block

Physics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

The correct option is D

Explanation:

This question is incomplete because of the absence of the setup which as been attached below. The setup shows/determines/tests the friction of wood (which is a block material), since Jerry wants to test the friction between different types of materials, he will have to replace the wooden block with another type of block material of choice so as to determine the friction of that also.

In order to have a comprehensive experiment, Jerry can use 4-5 different types of block material in the course of the experiment.

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A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³​
Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

                                                = 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

                        9.5 N = 10 N - weight of water displaced

                       Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

                                                            = 0.051 kg

Therefore the volume of the water displaced,

                                              V = m / ρ

                                                  = 0.051 / 1000

                                                   = 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

                                                                             = 5.1 x 10⁻⁵ x 800

                                                                             = 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

                                                                    = 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

                             W₀ = W - w

                                    = 10 N - 0.4 N

                                     = 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0
3 years ago
In an experiment to determine the s.h.c. of lead, a 0.80 kg block of lead is heated using a
gladu [14]

Answer:2.47

Explanation: did the math

8 0
2 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
A sound wave is called a longitudinal wave? why?Give reason
Mars2501 [29]

Answer:

A sound wave is called a longitudinal wave because compressions and rarefactions in the air produce it. The air particles vibrate parallel to the direction of propagation.

3 0
3 years ago
An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object
tatiyna

Answer:

D_T=18.567m

Explanation:

From the question we are told that:

Acceleration a=8.0 m/s^2

Displacement d=1.05 m

Initial time t_1=6.0s

Final Time t_2=2.5s

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

 V^2=2as

 V=\sqrt{2*6*1.05}

 V=4.1m/s

Generally the equation for Distance traveled before stop is mathematically given by

 d_2=v*t_1

 d_2=3.098*4

 d_2=12.392

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity v_3=0 m/s

Initial velocity u_3=4.1 m/s

Therefore

Using Newton's Law of Motion

 -a_3=(4.1)/(2.5)

 -a_3=1.64m/s^2

Giving

 v_3^2=u^2-2ad_3

Therefore

 d_3=\frac{u_3^2}{2ad_3}

 d_3=\frac{4.1^2}{2*1.64}

 d_3=5.125m

Generally the Total Distance Traveled is mathematically given by

 D_T=d_1+d_2+d_3

 D_T=5.125m+12.392+1.05 m

 D_T=18.567m

6 0
3 years ago
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