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3 years ago

5

1. List a similarity between magnetic force and electrical force.

1 answer:

Rus_ich [418]3 years ago

7 0

Answer:

Both are attractive as well as repulsive.

Explanation:

(Like poles repel, like charges<em> repel</em>; unlike poles attract, unlike charges <em>attract</em>).

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What is number 30,31 in this picture?

Maslowich

On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C

Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)

on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.

Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:

C, B, A and then D

the same idea as on Q30 applies to Q31 part b,

D,C,B then A

6 0

3 years ago

A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

Mazyrski [523]

Answer:

Coefficient of friction is $0.068$ .

Work done is $320~J$ .

Explanation:

Given:

Mass of the box ( $m$ ): $60$ kg

Force needed ( $F$ ): $40$ N

The formula to calculate the coefficient of friction between the floor and the box is given by

$F=\mu mg...................(1)$

Here, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Substitute $40$ N for $F$ , $60$ kg for $m$ and $9.80$ m/s² for $g$ into equation (1) and solve to calculate the value of the coefficient of friction.

$40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068$

The formula to calculate the work done in overcoming the friction is given by

$W=Fd..........................(2)$

Here, $W$ is the work done and $d$ is the distance travelled.

Substitute $40$ N for $F$ and $8 m$ for $d$ into equation (2) to calculate the work done.

$W=40~N\times8~m\\~~~~= 320~J$

8 0

3 years ago

A 3.80-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building

Virty [35]

Answer:

Explanation:

The question is ;

3.80 m-long, 500. kg steel beam extends horizontally from the point

where it has been bolted to the framework of a new building under

construction. A 67.0 kg construction worker stands at the far end of

the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

Solution:

Here two torques are in action

a) torque T1 due to weight of worker at the edge of the beam - at center of the beam

b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted

So we first calculate the torque produced due to weight of the worker;

We can see that;

Distance of worker from the center of the beam = 1.9m

Mass of the worker = 67 kg

Value of g= 9.8 m/sec2

T1=force × distance from point of rotation

Here force is weight of the worker which is = mass × g=67×9.8=656.6 N

So the torque is

T1= 656.6×1.9=12225.892 Nm

or

T1 = 12225.892 Nm

Now torque by the beam itself ;

Length of the beam from its center point to the bolt point = 1.9 m

Weight force of beam acting at center point= mass× g= 500×9.8=4900 N

Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm

T2=9310Nm

So total torque = T1+T2= 12225.892+9310=21565.892 Nm

8 0

3 years ago

What would cause the pressure in a sealed container of gas to increase?

Whitepunk [10]

<span>a. The temperature increases.</span>

3 0

4 years ago

A 50 cm tall object is 3.0 m from a plane mirror. How tall will<br> the image be?

anastassius [24]

Answer: 3.0 m

Explanation:

5 0

3 years ago