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3 years ago

11

Problem 3.2 Part A Draw the vector C⃗ =A⃗ +B⃗ .(Figure 1)

1 answer:

Elodia [21]3 years ago

5 0

I can't answer this question without a figure. I've found a similar problem as shown in the first picture attached. When adding vectors, you don't have to add the magnitudes only, because vectors also have to factor in the directions. To find the resultant vector C, connect the end tails of the individual vectors. <em>The red line (second picture) represents the vector C.</em>

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A man, facing a high wall, notices that an echo is heard 4.0 seconds after he makes a sharp sound. After walking 200m directly t

cestrela7 [59]

Answer:2.4

Explanation:basically 2.4 = 2.500 kg and then we add 3.4 seconds in all of them 4.0 + 3.4 = 2.5 ok

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3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

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3 years ago

Give v=120m/s north and t = 10.2, calculate the displacement

Sedbober [7]

Answer:

Displacement from the starting position is 1224m

Explanation:

The object is moving at 120 meters per second, it does this for 10.2 seconds. Use multiplication to find the answer.

$120 \times 10.2 = 1224$

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3 years ago

A 185 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring

denis-greek [22]

Answer:

d = 5.10 m

Explanation:

As we know that here on the plane of the inclined there is no frictional force

So in these cases we can say that total mechanical energy will always remains conserved

so here we can say that

spring potential energy = gravitational potential energy of the block

as we know from the formula

$\frac{1}{2}kx^2 = mgh$

now plug in the values in it

$\frac{1}{2}(1.60 \times 10^3)(0.10)^2 = (0.185)(9.81)h$

$8 = 1.81 h$

$h = 4.42 m$

now as we know that the angle of inclination is 60 degree and height raised is 4.42 m

so here maximum distance moved along the inclined plane will be

$\frac{h}{d} = sin60$

$d = \frac{h}{sin60}$

$d = \frac{4.42}{sin60} = 5.10 m$

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3 years ago

Visible light transmitting through a certain medium.

Verizon [17]

I think its A: Translucent hope this helps if I'm wrong I'm sorry

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3 years ago