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asambeis [7]
3 years ago
13

A uniform brick of length 21 m is placed over

Physics
1 answer:
Paha777 [63]3 years ago
6 0

Answer:

15.75 m

Explanation:

First, let's look at the top brick by itself.  In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick.  So the edge of the top brick must be 10.5 m from the edge of the bottom brick.

Now let's look at both bricks as a combined mass.  We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m.  And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge.  So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

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Which term is a scalar quantity?
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Vector quantities have both magnitude and direction. Distance is a scalar quantity. It refers only to how far an object has traveled. For example, 4 feet is a distance; it gives no information about direction.
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3 years ago
Read 2 more answers
True or False: Amplitude can increase or decrease wavelength
Marat540 [252]

Answer:

A. False, frequency can increase or decrease wavelength.

For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.

4 0
2 years ago
The answer and how to do it
Arturiano [62]
Current = charge per second
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I think you can find the answer now.
8 0
3 years ago
I need help on number 5. We need to use one of the four kinematics equations.
Gnom [1K]

ANSWER

35.02m

EXPLANATION

Parameters given:

Initial velocity, u = 26.2 m/s

When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.

We can now apply one of Newton's equations of motion to find the height:

v^2=u^2-2as

where a = g = acceleration due to gravity = 9.8 m/s²

Therefore, we have that:

\begin{gathered} 0=26.2^2-2(9.8)s \\ \Rightarrow19.6s=686.44 \\ s=\frac{686.44}{19.6} \\ s=35.02m \end{gathered}

That is the height that the vase will reach.

5 0
7 months ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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