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Tju [1.3M]
3 years ago
14

100 J of work in 10 seconds

Physics
1 answer:
HACTEHA [7]3 years ago
7 0
The average power is 10 watts. What's the question ?
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Which of the following describes an accelerated motion
Naily [24]
Accelerated motion such as a vehicle (car) or a moving item such as a (football thrown in the air)
3 0
3 years ago
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

4 0
3 years ago
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.
Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

=\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}

a = 30 units    and  b = 70 units

= \sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}

=   40 units to 100 units

hence, the correct answer is option C which is 50 units.

                       

4 0
3 years ago
Arunning back practices for his upcoming football game by running drills. He runs forward 62.4 yards, and runs backward for
Mama L [17]

Answer: He has only move 0.2 yards

Explanation: When you subtract 18.3 from 18.5 you get 0.2 and that is how much he's moved

4 0
3 years ago
.
beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s)  = 9m

Time (t)  = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0
3 years ago
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