Light travels in...<br> A) electromagnetic waves B) electrons <br> C) photons <br> D) a and c

a car slows down from 22 m/s to 3 m/s at a constant rate of 3.5 m/s squared. how many seconds are required before the car is tra

Stolb23 [73]

Answer:

9.05 s

Explanation

v=u+at

6 0

3 years ago

A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

kirza4 [7]

Answer: $v=\sqrt{\frac{FL}{m}}$

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

$F=T=\frac{mv^2}{L}$

$v=\sqrt{\frac{FL}{m}}$

8 0

3 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

A 50 kg astronaut floating in space throws her 2 kg wrench to the left at 10

Hoochie [10]

The astronaut will move at 0.4 m/s in the opposite direction to the wrench

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the astronaut-wrench system must be conserved before and after the launch.

Before the launch, the total momentum is zero, since the astronaut is at rest:

p = 0 (1)

After the launch, the total momentum is:

$p=mv+MV$ (2)