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2 years ago

How can you cause a cart moving to the right at a constant speed to move to the left at a constant speed.

Physics

1 answer:

lozanna [386]2 years ago

6 0

Answer:

The answer is below hoped it helped

Explanation:

Forces, when unbalanced, cause objects to accelerate. And the direction of the acceleration is the same as the direction of the extra force. Just because an object is moving to the right does not mean there is more right force than left force. Whether the right or the left force is bigger depends on whether the object is speeding up or slowing down. There must be more right force than left force if an object is moving to the right and speeding up. There must be more left force than right force if an object is moving to the right and slowing down.

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Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R

vodomira [7]

Answer:

$V = IR$

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

$I\ \alpha\ \frac{V}{R}$

Convert the expression to an equation

$I\ =\ \frac{V}{R}$

Multiply both sides by R to make V the subject

$I\ * R\ =\ \frac{V}{R} * R$

$I\ * R\ =V$

Reorder

$V = I\ * R$

$V = IR$

<em>Option (a) is correct; Others are not</em>

6 0

2 years ago

Read 2 more answers

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago

A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph

Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0

3 years ago

Yes or no

Tju [1.3M]

Answer:

1. No, it is a colloid

2. Yes

Explanation:

8 0

2 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0

3 years ago