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sashaice [31]
2 years ago
15

A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string

horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80 kg block that is at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3 m/s. What is the speed of the ball just after the collision
Physics
1 answer:
kupik [55]2 years ago
6 0

Answer:

the speed of the ball just after the collision is 1.5 m/s.

Explanation:

Given;

mass of the ball, m₁ = 1.6 kg

initial velocity of the ball, u₁ = 0

mass of the block, m₂ = 0.8 kg

initial velocity of the block, u₂ = 0

final velocity of the block, v₂ = 3 m/s

let the final velocity of the ball after collision = v₁

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.6 x 0   +    0.8 x 0       =   1.6 x v₁     +  0.8 x 3

0 = 1.6v₁  + 2.4

-1.6v₁ = 2.4

v₁  = -2.4 / 1.6

v₁ = - 1.5 m/s

v₁ = 1.5 m/s (in opposite direction of the block)

Therefore, the speed of the ball just after the collision is 1.5 m/s.

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What is the contour interval of this map? a. 20 feet b. -20 feet c. 60 feet 11​
timofeeve [1]

Answer:

c. 60 feet is the correct answer

Explanation:

what is the contour interval of this map? a.20 b.-20 c. 60 feet 11

8 0
3 years ago
At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?
notka56 [123]

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

3 0
1 year ago
Anyone knows this? Please answer... Spam will be reported.
Yakvenalex [24]

Answer:

The correct option is;

The assertion is correct, but reason wrong

Explanation:

The question is with regards to the relationship between work, energy, power, and velocity

The mass of each of the persons running up the staircase = Different

The time it takes each person to run up the stairs = Equal time

Let, 'm₁' and 'm₂' represent the mass of each of the persons that ran up the stairs and m₁ > m₂

Let 't' represent the equal time it takes then to run up the stairs

Let 'h' represent the height of the stairs

The energy, 'E', it takes to run up the stairs is equal to the potential energy, P.E., obtained at the top of the stairs

P.E. = m·g·h

Where;

m = The mass of the person at an elevated height

g = The acceleration due to gravity = Constant

h = The height reached above ground level

Given that the height reached is the same for both of the persons, we have

For m₁, P.E.₁ = m₁·g·h and for m₂, P.E.₂ = m₂·g·h

Therefore, where, m₁ > m₂, we have;

P.E.₁ > P.E.₂

∴ E₁ > E₂

Power, 'P', is the rate at which energy is expended

∴ Power, P = E/t

∴ P₁ = E₁/t  > P₂ = E₂/t

Therefore, the person with the greater mass, 'm₁', uses more power than the person of mass 'm₂', in running up the stairs

Therefore, the assertion is correct

The average velocity, vₐ = (Total distance traveled, d)/(Total time taken, t)

Given that the distance, 'd', covered in running up the stairs by both persons is the same, and the time it takes them to complete the distance, 't', is also the same, we have;

The average velocity of the person with the greater mass m₁ is the same as the average velocity of the person with mass, m₂

Therefore, the reason is wrong

The answer is that the assertion is correct, but reason wrong

6 0
2 years ago
The answer and how to do it?? Thanks
denis-greek [22]

Answer:

14 m/s²

Explanation:

Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.

(4N + 10 x 1kg)=(1kg)a

14/1=14, so the acceleration is 14 m/s²

4 0
3 years ago
A quantity of a gas is an absolute pressure of 400 K PA in the absolute temperature of 110° Kelvin when the temperature of the g
aliina [53]

Given:

P1 = 400 kPa
T1 = 110 K

T2 = 235K

Required:

P2

Solution:

Apply Gay-Lussac’s law where P/T = constant

P1/T1 = P2/T2

P2 = T2P1/T1

P2 = (235K)(400kPa) / (110K)

P2 = 855 kPa

4 0
3 years ago
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