A mass on a string is swung in a circle of radius 0.75m at 7.0m/s.what its rate of acceleration.

Explain why friction makes it harder to push a box filled with groceries than an empty box

kupik [55]

It is harder to push a box with groceries in it because the mass is weighing the box down. Which makes it harder to push

5 0

3 years ago

A person is in a car crash and is not wearing their seatbelt. They hit their

Greeley [361]

He still move deforce stopping this call law of inertia depend on the mass of object

5 0

3 years ago

Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you

Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

= 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t

v = u+at

t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min

t = 27.2 min

4 0

3 years ago

A man supports himself and the uniform horizontal beam pulling the rope with a force T.The weights of men and the beam are 883 N

artcher [175]

Answer:

T=502.5N

Ax=171.8N

Explanation:

The computation of the tension T in the rope and the forces exerted by the pin at A is shown below:

vertical forces sum = Ay + Tsin20 + T - 245 - 883 = 0

Now

horizontal forces sum = Ax - Tcos70

Now Moment about B

-Ay × 4.8 + 245 × 2.4 + 883 × 1.8=0

Ay=453.6N

Now substitute in sum of vertical forces T=502.5N

Ax=171.8N

3 0

3 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

3 years ago

A mass on a string is swung in a circle of radius 0.75m at 7.0m/s.what its rate of acceleration.​

A mass on a string is swung in a circle of radius 0.75m at 7.0m/s.what its rate of acceleration.