All categories

2 years ago

X applies a force of 500 N to push Y 100 m across a road how much work does X do?

Physics

1 answer:

dolphi86 [110]2 years ago

7 0

Answer:

from the formula of work done =Force *Distance 500*100=50000

You might be interested in

in which type of wave are vibrations at right angles to the direction in which the wave is travelling

Pani-rosa [81]

Answer:

longitudinal waves have those properties

7 0

2 years ago

a crane lifts four pallets of bricks each of which weigh 5000 N. the crane lifts each pallet a height of 30m. the crane takes 4

love history [14]

Answer:

625 W

Explanation:

Applying

P = W/t.................... Equation 1

Where p = power, W = Work, t = time

But,

W = Force (F) × distance (d)

W = Fd........................ Equation 2

Substitute equation 2 into equation 1

P = Fd/t.................... Equation 3

From the question,

Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds

Substitute these values into equation 3

P = (5000×30)/240

P = 625 Watt

3 0

2 years ago

As part

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

2 years ago

A transformer has input voltage and current of 12 V and 4 A

Advocard [28]

Explanation:

It is given that,

Input voltage, $V_i=12\ V$

Input current, $I_i=4\ A$

Output current, $I_o=0.8\ A$

Number of turns in the secondary side of transformer, $N_s=1177$

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

$\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$

Substituting all the above values

So,

$N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885$

So, the number of turns in primary side of the transformer is 5885.

4 0

2 years ago

Choose the correct statement regarding a compound microscope being used to look at a mite. The eyepiece lens forms a virtual ima

Nata [24]

Answer:

a) True. The image of the mite is virtual

e) True. The image must be within the focal length of the eyepiece len

Explanation:

Let's review the general characteristics of compound microscopes

Formed by two converging lenses

Magnification is

M = -L/fo 0.25/fe

Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length

Let's review the claims

a) True. The image of the mite is virtual

b) False. The effect is the opposite of the magnification equation

c) False. The objective lens forms a real image

d) False. As the seal distance increases the magnification decreases

e) True. The image must be within the focal length of the eyepiece len

4 0

3 years ago

X applies a force of 500 N to push Y 100 m across a road how much work does X do?​

X applies a force of 500 N to push Y 100 m across a road how much work does X do?