Your shopping cart has a mass of 65 kilograms, In order to accelerate the shopping cart down an aisle at 0.3 m/sec2, what force would you need to use or apply to the cart?

Answers

Answer 1
Answer: remember Newton's second law: F=ma

to get the force in newtons, mass should be in kg and acceleration in m/s^2

conveniently, we don't need to convert units

we just need to multiply the two to get the force

65* 0.3 = 19.5 kg m/s^2 or N

if significant digit is an issue, the least number if sig figs is 1 so the answer would be 20 N
Answer 2
Answer:

Final answer:

To accelerate a shopping cart with a mass of 65kg at a rate of 0.3 m/sec2, you would need to exert a force of 19.5 Newtons, according to Newton's second law of motion.

Explanation:

The subject of this problem is physics, specifically the concept of force, mass, and acceleration within the domain of Newton's second law of motion. The law states that the force needed to accelerate an object is equal to the mass of the object multiplied by the desired acceleration.

Given the mass of the cart is 65 kilograms and the acceleration is 0.3 m/sec2, you can calculate the required force using the formula:
Force = mass * acceleration.

So, Force = 65 kg * 0.3 m/sec2 = 19.5 Newtons. Therefore, you would have to exert a force of 19.5 Newtons to accelerate the shopping cart at the specified rate.

Learn more about Force calculation here:

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What substance does rock behave like when it is exposed to extreme temperature and pressure?

Answers

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when an incandescent light bulb is turned on, its thin wire filament heats up quickly. as the temperature of this wire filament increases, its electrical resistance...

Answers


Well, we usually assume that the resistance of a circuit component
is constant and doesn't change.  But the truth is that for anything
that conducts current, its resistance always increases somewhat
when it warms up. 

For things like light bulbs, electric toasters, space heaters, electric
stove burners, the heat coils in a blow-dryer ... anything that's
designed to be really hot when it's doing its job ... the resistance
of those things increases significantly when they come up to their
operating temperatures.

Hello user!

here is it answer.....

Because the lamp filament is physically very small and dissipates a relatively large amount of power (say, 60W for a typical incandescent bulb), while the conductors which feed it are relatively large and of negligible resistance, so they dissipate very little power, for a relatively large conductor area. How hot a component gets from resistive heating is proportional to its power dissipation, and inversely proportional to its area/size.

Thus if you dissipate, say 60W of power in a tiny lamp filament, it will get hot enough to become (surprise !!) incandescent, while the conductors which supply it (which dissipate perhaps a fraction of a watt for many feet of conductor length) will experience no significant temperature rise.

In fact, applicable electrical codes mandate that circuit conductors be sized large enough and provided with appropriate overcurrent protection, so that no significant conductor heating can take place.

. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0degrees above the horizontal from the top of a cliff 49.0 m high.
Calculate:
(a) the time to reach maximum height
(b) the maximum height above the base of the cliff reached by the
projectile
(c) thetotal time it is in the air
(d) the horizontal range of the projectile.​

Answers

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m.  Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m.  Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

Which force changes Earth's surface most slowly?A) earthquake
B) erosion
C) volcano
D) landslide

Answers

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^_^...hope this helps!!!!

As the bell rings, which energy transformation occurs? A. Mechanical energy is converted into sound energy. B. Chemical energy is converted into heat energy. C. Mechanical energy is converted into heat energy. D. Chemical energy is converted into sound energy.

Answers

Answer:

A) Mechanical energy is converted into sound energy.

Explanation:

Answer:

Mechanical Energy.

Explanation: