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11 days ago

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A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displa

cement, the work it does is equal to how many joules?

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If you start with the number 3.0 and move the decimal point one unit to the left, you wind up with 0.30. If you move the decimal

Keith_Richards [3271]

Answer:

10000

Explanation:

When shifting the decimal point left, the number effectively gets divided by 10 for each movement.

Conversely, shifting it to the right corresponds to multiplying the number by 10 for every unit moved.

If the decimal point is moved four units to the left, it becomes 0.0003, which is equivalent to dividing 3.0 by 10000

$\Rightarrow \frac{3.0}{10000} = 0.0003$

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3 months ago

A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.

inna [3103]

Answer:

44.4m/s^2

Explanation:

Utilize the equation...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

; The acceleration due to gravity is 44.4m/s^2

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3 months ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

5 0

3 months ago