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2 years ago

A cart is uniformly accelerating from rest. Which of the following best describes the net force acting on the cart?

Physics

2 answers:

eimsori [14]2 years ago

7 0

I think the answer is constant.

DerKrebs [107]2 years ago

3 0

Answer:

the answer is constant

Explanation:

the acceleration is uniformly accelerating which means it's at a constant.

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A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force

boyakko [2]

Answer:

Explanation:

7 0

1 year ago

Read 2 more answers

A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

$CPR=\frac{KW}{DAT}$

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6 to give CPR in mm/yr

mpy

$CPR=\frac{KW}{DAT}$

$=\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}$

=37.4mpy

mm/yr

$CPR=\frac{KW}{DAT}$

$=\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}$

=0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0

2 years ago

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50.

Nady [450]

Answer:

v = 38.73 m/s

Explanation:

Given

Extension of the bow, x = 50 cm = 0.5 m

Force of the arrow, F = 150 N

Mass of the arrow, m = 50 g = 0.05 kg

speed of arrow, v = ? m/s

We start by finding the spring constant

Remember, F = kx, so

k = F/x

k = 150 / 0.5

k = 300 N/m

the potential energy if the bow when pulled back is

E = 1/2kx²

E = 1/2 * 300 * 0.5²

E = 0.5 * 300 * 0.25

E = 37.5 J

The speed of the arrow will now be found by using the law of conservation of energy

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m, on substituting, we have

v² = (300 * 0.5²) / 0.05

v² = 75 / 0.05

v² = 1500

v = √1500

v = 38.73 m/s

8 0

2 years ago

The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the curre

Anestetic [448]

Answer:

energy carried by the current is given by the pointyng vector

Explanation:

The current is defined by

i = dQ / dt

this is the number of charges per unit area over time.

The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.

But the energy carried by the current is given by the pointyng vector of the electromagnetic wave

S = 1 / μ₀ EX B

It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement

5 0

2 years ago

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal

Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

p = m1 v1 + m2 v2

Part B.

Find the derivative

dp / dt = m1 dv1dt + m2 dv2 / dt

dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1 f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0

2 years ago