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Maksim231197 [3]
3 years ago
7

John doe gets on the highway in his 1967 Shelby 427 Cobra starting from the dead stop at the bottom of the on ramp of it can be

assumed that John accelerates at a rate of 6.7 m s2 how long does it take for John to reach the speed of 30 m s?
Physics
1 answer:
masha68 [24]3 years ago
7 0

Answer:

It will take john 4.477 seconds

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A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to
geniusboy [140]

Answer:

4.3 km

Explanation:

5 0
2 years ago
I've been on this for a while please help me
True [87]

Answer: Student 2

Explanation: Iron nail and a paperclip are conductors because they are made of metal. A rock, rubber band, and wooden stick are insulators because they cannot conduct electricity.

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2 years ago
The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The
Umnica [9.8K]

Answer:

v_{rms} =196.59 m/s

Given:

Temperature, T = 3.13 K

molar mass of molecular hydrogen, m = 2.02 g/mol = 2.02\times 10^{-3}kg/mol

Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

v_{rms} = \sqrt{\frac{3TR}{m}}

where

R = rydberg's constant = 8.314 J/mol-K

Putting the values in the above formula:

v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}

v_{rms} =196.59 m/s

5 0
3 years ago
What is the scientific definition of tracked?
dexar [7]
To find or discover by investigation?
8 0
3 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
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