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lapo4ka [179]
2 years ago
5

Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity

Physics
1 answer:
AfilCa [17]2 years ago
5 0

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ4

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one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

   (150 horsepower) x (746 watt/HP) x (1 joule/sec  /  watt) x (10 sec)

=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
2 years ago
provides one model for solving this problem. The maximum strength of the earth's magnetic field is about 6.9 x 10-5 T near the s
jeka57 [31]

Answer:

The minimum no. of turns is 3.126 \times 10^{5}

Explanation:

Given:

Magnetic field B = 6.9 \times 10^{-5} T

Frequency f = 84.5 Hz

Area of turn A = 0.021 m^{2}

Voltage V_{rms}  = 170 V

From the formula of induced emf,

V = NBA \omega

Where \omega = 2\pi f and V = \sqrt{2} V_{rms}

So number of turn is,

N = \frac{\sqrt{2} V_{rms} }{AB2\pi f }

N = \frac{\sqrt{2} \times 170 }{0.021 \times 6.9 \times 10^{-5} \times 6.28 \times 84.5 }

N = 3.126 \times 10^{5}

Therefore, the minimum no. of turns is 3.126 \times 10^{5}

4 0
3 years ago
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
3 years ago
A tennis ball is thrown straight up into the air with an initial velocity of 38 m/s.
geniusboy [140]

Answer:

a) 3.9s

b)74m

Explanation:

a) v=u+at

0=38+(-9.8)t

-38=-9.8t

t=-38/-9.8

t=3.877s

t=3.9s

b) s=ut+1/2at²

s=(38×3.9)+1/2(-9.8×3.9²)

s=148.2 -74.529

s=73.671

s=74m

8 0
2 years ago
How does the electric force between two charged particles change if one particle's charge is increased by a factor of 2?
Burka [1]
 if one of the charge is doubled, the electric potential energy would be doubled too
8 0
3 years ago
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