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3 years ago

A worker pushes horizontally on a large crate with a force of 170 N, and the crate is moved 3.6 m. How much work was done

Physics

1 answer:

Mariulka [41]3 years ago

8 0

W = F * D

W = 170 N * 3.6 M

W = 612 J

You can use this format to answer questions where you're given the force and distance.

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What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k

Molodets [167]

Answer:

$\dfrac{I_1}{I_2}=12.25$

Explanation:

$r_1$ = 14 km

$r_2$ = 49 km

Intensity of a wave is inversely proportional to distance

$I\propto \dfrac{1}{r^2}$

So,

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25$

The ratio of the intensities is $\dfrac{I_1}{I_2}=12.25$

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3 years ago

The sound level near a noisy air conditioner is 70 dB. If two such units operate side by side, the sound level near them would b

Schach [20]

Answer:

73 db

Explanation:

A single air conditioner is equivalent to 70 dB frequency. An extra air conditioning unit would therefore double the sound frequency. It does not, however, double the decibels to 140 dB. Instead, it adds only 3 dB to the 70 dB, making the total decibels of two air conditioning units equal to 73 db

Hence the correct option is b that is 73 db

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4 years ago

Differentiate between s-block and p-block

alexandr402 [8]

Explanation:

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2 years ago

Read 2 more answers

A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball

RideAnS [48]

Answer:

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, $F_r=Cv^2$

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

$Cv^2=mg$

$C=\dfrac{mg}{v^2}$

$C=\dfrac{0.046\times 9.8}{(44)^2}$

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Hence, this is the required solution.

4 0

3 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

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3 years ago