An object with high momentum _____

I). Mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work.

Answer:

false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = - Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

4 0

3 years ago

Infer the direction of the net force acting on a car as it slows down and turns right.

Kobotan [32]

Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.

6 0

2 years ago

Read 2 more answers

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

$x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s$

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) $\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s$

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

6 0

3 years ago

9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be

Akimi4 [234]

Answer:

525 Bq

Explanation:

The decay rate is directly proportional to the amount of radioisotope, so we can use the half-life equation:

A = A₀ (½)^(t / T)

A is the final amount

A₀ is the initial amount,

t is the time,

T is the half life

A = (8400 Bq) (½)^(18.0 min / 4.50 min)

A = (8400 Bq) (½)^4

A = (8400 Bq) (1/16)

A = 525 Bq

8 0

3 years ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

An object with high momentum ______.