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3 years ago

A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?

Physics

2 answers:

NeTakaya3 years ago

7 0

5.0*10^3 this is the answer. I know for a fact.

sweet-ann [11.9K]3 years ago

7 0

W=Fd

W= (8.35 X10^2N)(6.0m)

W= 5.0 X 10^3J

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How do the units of work and power compare?

ki77a [65]

Answer:

Option (d) is correct.

Explanation:

Work done is given by :

W = Fd, where F is force and d is displacement

Unit of work done :

The SI unit of force is Newton (N) and that of displacement is meter (m). So, the unit of work done is N-m. It is call Joule. It means that the unit of work done is Joule.

Power is given by rate at which the work is done. It is given by :

P = W/t, W is work done and t is time

Unit of power:

Unit of work is Joule (J) and that of time is second (s). It means that the unit of power is Watt and it is equal to Joule/second

Hence, the correct option is (d) "The unit for work is a joule. The unit for power is a watt, which is a joule per second".

7 0

2 years ago

Where does the USA get most of it's steel?

Troyanec [42]

China because it is cheaper then getting it from Pittsburgh

6 0

2 years ago

Read 2 more answers

Four long wires are each carrying 6.0 A. The wires are located

Firdavs [7]

Answer:

$B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

$B=\frac{\mu_oI}{2\pi r}$

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

$B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]$

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

$r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m$

Then you have:

$B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

5 0

3 years ago

A 50 kg bicyclist, traveling at a speed of 12 m/s, applies the brakes, slowing her speed to 3 m/s.

Bezzdna [24]

a) Work done = Net Kinetic Energy

= 1/2 x 50 kg x ((12m/s)^2 - (3m/s)^2)

= 0.5 x 50 Kg x (144 -9)(m/s)^2

= 3375 Kg (m/s)^2

b) Force = mxa

a = 120 N/50 Kg = 2.4 m/s^2

Using newtons third law of motion, we get-

V^2 - U^2 = 2 x a x S

S= (12^2-3^2)m^2/s^2/(2 x 2.4 m/s^2)

= 28.125 m

3 0

3 years ago

1.) A negative charge of -2.0x10-4 C and a positive charge of 8.0x10-4 C are

exis [7]

Answer:

The Force between the two charges is an attractive force of 16,000N

Explanation:

Expression for the electric force between the two charges is given by

F = (k*q1*q2) / r^2

Here, k = constant = 9 x 10^9 N*m^2 / C^2

q1 = - 2.0x10^-4C

q2 = + 8.0x10^-4C

r = 0.30 m

Substitute the given values in the above expression -

One charge is + and the other is a -, therefore the net force is an attractive force (opposites atract)

The attraction force is:

F= 9.0x10^9 * 2.0x10^-4 *8.0x10^-4 N/ 0.30^2

F= 16,000N

6 0

2 years ago