A 10 kg ball has a momentum of 40 kg.m/s.what is the ball's speed?

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

now, calculating at the angular velocity

$\omega =\dfrac{2\pi}{T}$

$\omega =\dfrac{2\pi}{1.25}$

ω = 5.026 rad/s

now, radial acceleration

a = r ω²

a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0

3 years ago

3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final

MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m = 9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0

3 years ago

A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will

Aleks [24]

Answer:

We know that for a pendulum of length L, the period (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

7 0

3 years ago

catapult’s lever holds a cannonball. The lever is attached to a tightly held rope. When the rope is released, the lever springs

Shalnov [3]

Answer:

Potential energy

Explanation:

Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.

6 0

3 years ago

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

Gnom [1K]

Answer:

(a) Acceleration of positron is 6.03 x 10¹³ m/s²

(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s

Explanation:

Given :

Constant electric field, E = 343 N/C

Mass of positron, m = 9.1 x 10⁻³¹ kg

Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C

(a) Coulomb force on the positron is determine by the relation :

F = q x E ....(1)

But, force is also equals to product of mass and acceleration. So,

F = ma .....(2)

Here a is acceleration.

From equation (1) and (2).

m x a = q x E

$a=\frac{qE}{m}$

Substitute the values of q, E and m in the above equation.

$a=\frac{1.6\times10^{-19}\times 343}{9.1\times10^{-31} }$

a = 6.03 x 10¹³ m/s²

(b) Initially, the positron is at rest, so its initial speed is zero.

The equation of motion for positron is :

v = u + at

Here v is final speed, u is initial speed and t is time.

Since, u is zero, so the equation becomes :

v = at

Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.

v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s

v = 5.24 x 10⁵ m/s

6 0

3 years ago