You are in a car driving way from a stationary whistle. How does the sound you hear compare with that of the whistle itself?

tresset_1 [31]

Different forces can have simillar magnitudes and same directions (but opposite), in this case the result will be nothing would happen. Or one of forces can be bigger than the other/s (in the same direction) so the system would accelerate (F=ma). Or there could be a situation with the same magnitudes but different directions, so the object (system) would accelerate in the direction of secant line (F=ma)

6 0

3 years ago

According to Faraday's law, voltage can be changed by moving magnets away from the coil of wire. True False

Rasek [7]

The statement: “According to Faraday's law, voltage can be changed by moving magnets away from the coil of wire.”, is true. Any type of movements done (away, into or out, toward, rotation, etc.) on the magnetic environment would always generate a voltage. In Faraday’s law, it is clear that there are different ways on how to generate a voltage.

5 0

3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

ikadub [295]

Answer:

b) No acceleration in the vertical

c) 35N

d) 35N

e) $8.75\ m/sec^2$

Explanation:

a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book

b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero

c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

$Fh_{net}=F-F_r=50 N - 15N=35N$

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:

$F_n=35N$ in the x-axis

e) Following Newton's second law, the acceleration is calculated as

$a=\frac{F_t}{m}=\frac{35N}{4Kg}=8.75m/sec^2$

8 0

3 years ago

You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) how much work do you do on the can of pa

MariettaO [177]

(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

$W=mg\Delta h$

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and $\Delta h=1.8 m$ is the variation of height. Substituting the numbers into the formula, we find

$W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J$

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

$W=Fd$

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.

(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

$W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J$

8 0

3 years ago

A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

Flauer [41]

<h3>It takes 60 seconds to do the work</h3>

Solution:

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

Find the work done:

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

Find the time taken

$power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second$

Thus it takes 60 seconds to do the work

3 0

3 years ago