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kvasek [131]
3 years ago
14

There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

l block is 25 kg, and the large block is 50 kg. The blocks are to the left of a hill.The hill is 10 meters tall. (Drawing not to scale)a) The small block collides with the large block and sticks together. How fast must the initial velocityof the small block be so that the two blocks just reach the top of the hill. (Assume no friction).b) What was the total impulse the small block exerted on the large block in order to get it going fastenough
Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2  x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75  x 42 / 3

= 1050 Ns.

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
1. Calculate the average velocity of the following trip. You walk to Pershing Square 58
Gre4nikov [31]

Explanation:

Velocity = displacement / time

v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)

v = 0.20 m/s

7 0
3 years ago
Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn
mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

5 0
3 years ago
A 25 n object requires a 5.0 n to start moving over a horizontal surface. what is the coefficient of static friction?
olganol [36]
Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N =  0.2
Hence 0.2 is the coefficient of static friction
3 0
3 years ago
Read 2 more answers
A skater has outstretched arms preparing for a turn with a moment of inertia of
Helen [10]

Answer:

The final angular speed is 16.1 rad/s

Explanation:

Given;

initial moment of inertia, I₁ = 2.56 kg.m²

final moment of inertia, I₂ = 0.40 kg.m²

initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s

Apply the principle of conservation of angular momentum;

I₁ω₁ = I₂ω₂

where;

ω₂ is the final angular speed

ω₂ = (I₁ω₁) / (I₂)

ω₂ = (2.56 x 2.514) / (0.4)

ω₂ = 16.1 rad/s

Therefore, the final angular speed is 16.1 rad/s

7 0
3 years ago
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