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3 years ago

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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using mass and distance, identify and compare the sun's and moon's contribution to the formation of tides on earth

vekshin1

Answer:

Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.

7 0

2 years ago

You do 45.0 joules of work and 3.00 seconds how much power do you use

Arada [10]

Answer:

15 watt

Explanation:

Power is the rate at which work is done.

This means you divide the work done with the amount of time used to perform the work.

The formula for Power is : P = W/t where;

W= work done in J = 45

t= time in seconds = 3 sec

P= 45/ 3 = 15 watt

8 0

2 years ago

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le

Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

$R = u² sin2θ/g$

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

$u² = \frac{Rg}{sin2\theta}$

$u^2 = \frac{9.1 x 9.80}{sin26}$

$u^2 = 113.17$

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

$v^2 = (v_osin\theta)^2 -2gh$

$h =\frac{(v_o^2sin\theta)^2}{2g}$

$h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}$

h = 1.10 m

7 0

2 years ago

When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point

vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

$v = Velocity \rightarrow 20m/s$

$\lambda = Wavelength \rightarrow 35*10^{-2}m$

Therefore the frequency would be given as

$f = \frac{20}{35*10^{-2}}$

$f = 57.14Hz$

The frequency is directly proportional to the angular velocity therefore

$\omega = 2\pi f$

$\omega = 2\pi *57.14$

$\omega = 359.03rad/s$

Now the maximum speed from the simple harmonic movement is given by

$V_{max} = A\omega$

Where

A = Amplitude

Then replacing,

$V_{max} = (1*10^{-2})(359.03)$

$V_{max} = 3.59m/s$

Therefore the maximum speed of a point on the string is 3.59m/s

8 0

2 years ago

Which is a characteristics of a physical change?

vazorg [7]

Physical Change characteristic is the chemical bonds in the substance are unchanged. Because a physical change is any change happens in an object but without involving a change in its chemical substance. Example, Solid to liquid change or also known as melting, liquid to gas change also known as evaporation, gas to solid change also known as deposition, liquid to solid or solidification, solid to gas or sublimation, and gas to liquid or condensation. The physical form of a substance is change into a new form but the chemical is unchanged.

8 0

2 years ago