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7nadin3 [17]
3 years ago
8

6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar

Physics
1 answer:
Novay_Z [31]3 years ago
6 0

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

#LearnWithEXO

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10 points
enot [183]

Answer:

B

Explanation:

I just took the test

8 0
3 years ago
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D
RideAnS [48]

Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² =  vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M  vi² + FD) / 1/2 M  vi²

RK = 1 + FD / 2 M  vi²

4 0
3 years ago
Which of the following can only be a situation of increasing temperature?
Papessa [141]

Increasing the temperature causes an increase in the average kinetic energy of the particles of a material.

<h3>What is average kinetic energy of particles?</h3>

The average kinetic energy of particles is the energy possessed by particles due to their constant motion.

The constant motion of particles occurs due to the energy acquired by the particles, when the temperature of the particles increases, the average kinetic energy increases which in turn increases the speed of the particles.

Thus, we can conclude that, increasing the temperature causes an increase in the average kinetic energy of the particles of a material.

Learn more about average kinetic energy here: brainly.com/question/9078768

5 0
2 years ago
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
2 years ago
In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second
lakkis [162]
If he keeps that pace he will be at the 34 yard line
7 0
3 years ago
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