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3 years ago

Does an increase in velocity necessarily mean an increase in acceleration?

Physics

2 answers:

lilavasa [31]3 years ago

5 0

Yes. When Velocity increases acceleration does too.

Vinil7 [7]3 years ago

3 0

We know, acceleration = final velocity - initial velocity / time
Here, if velocity is increasing, then,
Final velocity > initial velocity, in that case, acceleration is also increasing, as it is directly proportional to velocity

In short, Your Answer would be "Yes"

Hope this helps!

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Martha (50 kg) is attracted to Stewart (70 Kg) who sits 4 m away. What is the gravitational attraction between them? G=6.67 x 10

kodGreya [7K]

Answer:

1.46× 10^-8 N

Explanation:

F=Gm₁m₂/r²

F=(6.67×10⁻¹¹ × 50× 70)/(4)²

F=1.46× 10^-8 N

Hope, this helps you.

8 0

3 years ago

A 3.00-kg object has a velocity 16.00 i ^ 2 2.00 j ^2 m/s.

trapecia [35]

Answer:

390 J

Explanation:

m = 3 kg

u = 16 i + 2 j

(a) Magnitude of velocity = $\sqrt{16^{2}+2^{2}}$ = 16.1245 m/s

KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J

(b) v = 18 i + 14 j

Magnitude of velocity = $\sqrt{18^{2}+14^{2}}$ = 22.804 m/s

KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J

According to the work energy theorem

Work done = change in KE = KEf - KEi = 780 - 390 = 390 J

8 0

4 years ago

A ball of mass 24.1 g is attached to a cord of length 0.417 m and rotates in a vertical circle. What is the minimum speed the ba

Karo-lina-s [1.5K]

Answer:

the minimum speed that the ball must have so that the cord does not become slack is 2.02 m/s.

Explanation:

In order to avoid slack, the centripetal force of the ball must equal its weight at the top of the circle. Therefore,

F_c = F_g

m v² / r = m g

v² = g r

v = √[g r]

v = √[(9.8 m/s²)(0.417 m)]

v = 2.02 m/s

Therefore, the minimum speed that the ball must have so that the cord does not become slack is 2.02 m/s.

6 0

3 years ago

A 0.270 m radius, 510-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

vlada-n [284]

Answer:

0.35701 T

Explanation:

$B_i$ = Initial magnetic field

$B_f$ = Final magnetic field

$\phi$ = Magnetic flux

t = Time taken = 4.17 ms

N = Number of turns = 510

$\epsilon$ = Induced emf = 10000 V

r = Radius = 0.27 m

A = Area = $\pi r^2$

Induced emf is given by

$\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T$

The magnetic field strength needed is 0.35701 T

4 0

4 years ago

When does a object have the greatest kinetic energy?

Rufina [12.5K]

When Object is at zero height, and there is no potential energy possess by the object then it exerts Greatest Kinetic energy in it's whole Journey

Hope this helps!

8 0

4 years ago