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Vikentia [17]
3 years ago
13

A rectangular block has dimensions 2.9 cm x 3.5 cm x 10.0 cm. the mass of the block is 615.0 g. what is the volume and density o

f the block? (answer it with correct numbers of significant figures)
Physics
1 answer:
Goshia [24]3 years ago
5 0
Density is the ratio of the substance's mass to volume. It is an extensive property, meaning it does not depend on the amount of the substance. The ratio is always fixed per identity of a substance. The density is equal to:

V = 2.9 cm x 3.5 cm x 10.0 cm = 101.5 cm³
Density = 615 g/101.5 cm³ = 6.06 g/cm³
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A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40
Vinvika [58]

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

5 0
3 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
What kind of radiation is stopped by a piece of paper
Zina [86]
Three types of radioation - Alpha, Beta, Gamma. hope this helps
4 0
3 years ago
Read 2 more answers
Calculate the gravitational potential energy of a body of mass 40 kg at a vertical height of 10 m. ( g = 9.8 m/s2)
olganol [36]
Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)
8 0
3 years ago
A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The
Arada [10]

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

<h3>What will be the maximum acceleration of the truck to avoid tipping over?</h3>

The maximum acceleration is obtained by taking clockwise moments about the tipping point of rotation.

Clockwise moment = Anticlockwise moment

Ft * 1.58 m = F * 0.67 m

where

  • Ft is tipping force = mass * acceleration, a
  • F is weight = mass * acceleration due to gravity, g

m * a * 1.58 = m * 9.81 * 0.67

a = 4.15 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

In conclusion, the acceleration of the truck is found by taking moments about the tipping point.

Learn more about moments of forces at: brainly.com/question/27282169

#SPJ1

3 0
2 years ago
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