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Irina-Kira [14]

3 years ago

5

A box is sitting stationary on a ramp that is 42° to the horizontal. The box has a gravitational force of 112.1 N. What is the m

inimum amount of force that can be applied to the box to get the box to move?

2 answers:

ale4655 [162]3 years ago

5 0

Answer:

Answer is 80

Explanation:

I just took the tets

nordsb [41]3 years ago

4 0

Force of gravity along the inclined plane

$F_x = F_g sin\theta$

here we have

$F_g = 112.1 N$

$\theta = 42 degree$

now we have

$F_x = (112.1 N)sin42$

$F_x = 75 N$

so here we require 75 N force along the incline to move the box upwards

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

4 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$

b

$M = 6$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 200 \ kg$

The length of the small object from the rock is $d = 2 \ m$

The length of the small object from the branch $l = 12 \ m$

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The force exerted by the weight of the rock is mathematically evaluated as

$W = m_r * g$

substituting values

$W = 200 * 9.8$

$W = 1960 \ N$

So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

Here $\theta$ is very small so $cos\theta * l = l$

and $cos\theta * d = d$

Hence

$F * l - W * d = 0$

=> $F = \frac{W * d}{l}$

substituting values

$F = \frac{1960 * 2}{12}$

$F =326.7 \ N$

The mechanical advantage is mathematically evaluated as

$M = \frac{W}{F}$

substituting values

$M = \frac{1960}{326.7}$

$M = 6$

6 0

3 years ago

3. Sailors use a capstan (shown below) to raise and lower the anchor. It takes two sailors located 1 meter from the center to tu

mr_godi [17]

Answer:

B. 2 meters.

Explanation:

To rotate the capstan a certain amount of torque is required, and if each sailor applies a force $F$ at a distance $D$ from the center, then for two sailors the total torque will be

$\tau = 2FD\\$ ;

therefore, for one sailor to apply the same torque it must be that the torque $\tau_2$ he applies must be equal to the torque that the two sailors applied:

$FD_2 =2FD$

which gives

$D_2 = 2D$ .

and since $D = 1\:meter$ ,

$\boxed{D_2 = 2\: meters}$

which is choice B.

4 0

3 years ago

Which interactions are part of the greenhouse effect? Select three options.

Allisa [31]

<u>Out of the given options, the following interactions are part of the greenhouse effect, </u>

Gases in the atmosphere absorb heat
Earth’s surface radiates energy back into the atmosphere
Gases in the atmosphere radiate heat back to the surface

Answers: Options A, D and E

<u>Explanation: </u>

The greenhouse effect, basically a warming effect caused by the greenhouse gases such as Carbon-Di-oxide, Methane, nitrous oxides, water vapour etc. These gases usually trap the heat that Earth Absorbs by the Sun.

In the day time, the Earth absorbs the energy in the form of heat which is radiated by the Sun. In the evening, the process gets reversed and the Earth starts releasing that heat into the atmosphere.

Now, this heat gets absorbed by this gases before it leaves the Earth's atmosphere and gets trapped there only, resulting in the temperature raise of the Earth's environment.

So, the prime causes of the greenhouse effect remains as the heat radiation from the Sun, the absorption of that heat by the Earth surface and the further absorption of that heat produced by the greenhouse gases that present in the atmosphere.

5 0

3 years ago

Read 2 more answers

A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.

AlexFokin [52]

<h2><u>We have</u>,</h2>

Initial velocity (u) = 0 m/s
Time taken (t) = 2.9s
Acceleration due to gravity (g) = + 10 m/s² [Down]

<h2><u>To calculate</u>,</h2>

Final velocity (v)
Height (h)

<h2><u>Solution</u><u>,</u></h2>

→ v = u + gt

→ v = 0 + 10(2.9)

→ v = 29 m/s $\qquad$ … ( Ans )

And,

→ h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

→ h = 42.05 m $\qquad$ … ( Ans )

4 0

3 years ago