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Orlov [11]
2 years ago
10

If each pull-up requires 300 J and Ben does a pull-up in 2 seconds, what is his power? 150 watts 300 watts 600 watts 750 watts

Physics
2 answers:
leonid [27]2 years ago
6 0

Answer:

150 watts

Explanation:

300/2 = 150 watts

Fudgin [204]2 years ago
4 0

Answer:

150 watts

Explanation:

The explanation is in the picture

PLEASE MARK BRAINLIEST

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If the spring constant is doubled , what value does the period have for a mass on a spring?
zhuklara [117]

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}

So the correct answer is

D. The period would decrease by sqrt (2)

8 0
3 years ago
Read 2 more answers
A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
What is the average speed of a kangaroo that hops 60 m in 5 s ?
jok3333 [9.3K]
The formula for average speed is
v_{avg}= \frac{total distance}{total time}
So we can just substitute our data.
v_{avg}= \frac{60}{5}=12 \frac{m}{s} - its the result
6 0
2 years ago
Read 2 more answers
To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 107 J of energy. Suppose we tried to store th
tatiyna

Answer:

9000RPM

Explanation:

"Angular velocity" is directly related to kinetic energy, that is, the Kinetic energy equation would allow an approximation to the resolution investigated in the problem.

The equation for KE is given by:

KE = \frac{1}{2} lw ^ 2

Now, starting from there towards the <em>Angular equation of kinetic energy</em>, the moment of inertia (i) is used instead of mass (m), and angular velocity (w) instead of linear velocity (V)

That's how we get

KE_{Angular} = \frac{1}{2} Iw^2

calculating the inertia for a solid cylindrical disk, of

m = 400kg

r = 1.2 / 2 = 0.6m

I_{disk} = \frac{1}{2} mr^2 = (0.5) (400) (0.6)^2 = 72 kgm^2

We understand that the total kinetic energy is 3.2 * 10 ^ 7J, like this:

3.2*10^7 = \frac{1}{2} Iw^2 = (0.5) (72) w^2 = 36w^2w^2 = 3.2*10^7 / 36 = 0.0888*10^7 = 88.8*10^4

w = 9.43*10^2 = 943 rad / s

Thus,

943 rad / s ≈ 9000 rpm

6 0
3 years ago
A 16 kg mass suspended from a light spring is replaced by a 4 kg mass. What factor changes the frequency of the oscillation? (a)
AnnZ [28]

Answer:

Frequency change by a factor of 2.

(b) is correct option.

Explanation:

Given that,

Mass = 16 kg

Replaced mass = 4 kg

We need to calculate the frequency

Using formula of frequency  

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Put the value into the formula

f_{1}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{16}}

f_{1}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{4}}....(I)

f_{2}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{4}}

f_{2}=\dfrac{1}{2\pi}\dfrac{\sqrt{k}}{2}}...(II)

f_{2}=2f_{1}

Hence, Frequency change by a factor of 2.

5 0
3 years ago
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