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2 years ago

5

A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f

orce?

1 answer:

lys-0071 [83]2 years ago

7 0

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 1.94 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 6 N

d = 0.4 m

$\theta = 36 degree$

so we will have

$W = 6 \times 0.4 cos36$

$W = 1.94 J$

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FromTheMoon [43]

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2 years ago

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If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb

rusak2 [61]

We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
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If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

4 0

2 years ago

A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of 0.0125 kg ⋅ m2 and a radius of 1

dimaraw [331]

Answer:

the tension in the string is 5.59 N

Explanation:

Here ,

m_1 = 0.385 Kg

m_2 = 0.710 Kg

Using second law of motion ,

a = F_net / effective mass

a = (0.710- 0.385)×9.8/(0.710 + 0.385 + 0.0125/0.15^2)

a = 1.93 m/s^2

Now , let tension be T ,

then,

mg-T=ma

0.710×g - T = 0.710×1.93

T = 5.59 N

the tension in the string is 5.59 N

7 0

1 year ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0

2 years ago

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e) y = 0 + ½ g t²

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t = √(2 125 / 9.8)

t = 5.05 s

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2 years ago