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2 years ago

A force of 345 N is applied to a 70 kg scooter. What will the scooter's acceleration be?

Physics

1 answer:

Harrizon [31]2 years ago

4 0

Explanation:

here, force applied (N) = 345 N

or, mass of scooter (m) = 70 kg

now, acceleration (a) = ?

we know, force = mass × acceleration

or, 345 = 70 × acceleration

or, acceleration = 345/70

therefore, acceleration = 4.9 m/s2 (meter/second square)

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Where are you most likely to build up enough static charge to receive a

FromTheMoon [43]

You're most likely to build up enough static charge to receive a shock by walking around in a carpeted restaurant in the desert. (A)

Walking on carpet is the fastest way to accumulate charge, and the dry desert air prevents the charge from dribbling off of you and away.

When I walked on stones in the Sinai Desert, the dry wind with a little bit of sand or dust in it built up enough static charge on me that I got a shock every time I stood less than a foot away from my partner.

I had the same experience a few years later near Ouarzazate in the interior of Morocco.

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3 years ago

Read 2 more answers

The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh

stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, $P_{headlight}$ = 38 W

The power at which the kick starter operates in parallel, $P_{kick \ starter}$ = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/ $P_{headlight}$ = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ $P_{kick \ starter}$ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

$P_{headlight}$ = I² × R₁

∴ $P_{headlight}$ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, $P_{headlight, S}$ ≈ 0.134 W

$P_{kick \ starter}$ = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, $P_{kick \ starter, S}$ ≈ 2.123 W

The total power they will consume, $P_{Total}$ = $P_{headlight, S}$ + $P_{kick \ starter, S}$

Therefore;

$P_{Total}$ ≈ 0.134 W + 2.123 W = 2.257 W

4 0

3 years ago

A truck carrying a 65.4-kg crate accelerates from rest to [07]____________________ km/hr on a flat horizontal surface in 15.2 se

NikAS [45]

Answer:

The work done on the crate by the truck is 9472.54 J.

Explanation:

Given that,

Mass of the truck, m = 65.4 kg

Initially, the object is at rest, u = 0

Final speed of the truck, v = 61.3 km /h = 17.02 m/s

m/s

Time, t = 15.2 s

We need to find the work done on the crate by the truck. The work done is equal to the change in kinetic energy of the truck as :

$W=\Delta K$

$W=K_f-K_i$

$W=K_f-0$

$W=\dfrac{1}{2}mv^2$

$W=\dfrac{1}{2}\times 65.4\times (17.02)^2$

$W=9472.54\ J$

So, the work done on the crate by the truck is 9472.54 J. Hence, this is the required solution.

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3 years ago

Electricity does not commonly occur in nature except in the form of ( hydrocarbons, photosyntheis, lightning, thunder)

garik1379 [7]

Lightning i would think

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3 years ago

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arlik [135]

The correct answer is 12

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3 years ago