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vovikov84 [41]
2 years ago
12

Please help. I don’t understand this

Physics
1 answer:
skad [1K]2 years ago
5 0

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

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Calculate the power required of a 60-kg person who climbs a tree 5 meters high in 10 seconds.
Alina [70]

Answer:

Explanation:

Power = Energy/time

-Don't have energy so I'm gonna solve for it

Gravitational Potential Energy = mass x gravity x height

= 60 kg x 9.8 m/s2 x 5m

= 2940 J

Power = Energy/time

=2940 J/10 s

= 294 W

3 0
2 years ago
Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in
sattari [20]

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

8 0
3 years ago
Suppose that the half-life of an element is 1000 years. How many half-lives will it take before one-eighth of the original sampl
34kurt

Answer:

3

Explanation:

The half-life of a radioactive isotope is the time it takes for the mass of the sample to halve.

This can be rewritten as follows:

m(t) = m_0 (\frac{1}{2})^n

where

m(t) is the mass of the sample at time t

m0 is the original mass of the sample

n is the number of half-lives that passed

We see that if we take n=3, the amount of original sample left is

m(t) = m_0 (\frac{1}{2})^3 = m_0 (\frac{1}{8})

So 3 (3 half-lives) is the correct answer.

3 0
3 years ago
Read 2 more answers
A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.
aleksley [76]
The acceleration that the cheeseburger experienced is 20 m/s^2.
6 0
3 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
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