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Illusion [34]
3 years ago
9

Determine the mass of an object if it's moving at 5 m/s and has a momentum of 50 kg*m/s

Physics
1 answer:
Ganezh [65]3 years ago
8 0
I think the mass is 250 kg
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A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
34kurt

Answer:

The centripetal acceleration of the runner is 1.73\ m/s^2.

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

v is the velocity of runner

v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s

Centripetal acceleration,

a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2

So, the centripetal acceleration of the runner is 1.73\ m/s^2. Hence, this is the required solution.

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3 years ago
Who was the first head of the NASA space program was Sergei Korolov, but the Russians never identified him by name. What did the
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Korolev's name became known in the West. I hope this helps.

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4 years ago
What does the term "catastrophic event" mean to you
ozzi

Answer:

an event which took place which had such a significant impact.

Explanation:

6 0
3 years ago
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At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of banan
Elina [12.6K]

Answer:

2.67kg

Explanation:

The maximum velocity, v _ {max} of a body experiencing simple harmonic motion is given by equation (1);

v_{max}=\omega A............(1)

where \omega is the angular velocity and A is the amplitude.

The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);

\omega=\sqrt{\frac{k}{m}}.................(2)

where k is the force constant of the spring and m is the loaded mass.

We can make \omega the subject of formula in equation (1) as follows;

\omega=\frac{v_{max}}{A}.................(3)

We then combine equations (2) and (3) as follows;

\frac{v_{max}}{A}=\sqrt{\frac{k}{m}}.................(4)

According to the problem, the following  are given;

v_ {max }=1.92m/s\\A=0.21m\\k=223N/m

We then substitute these values into equation (4) and solve for the unknown mass m as follows;

\frac{1.92}{0.21}=\sqrt{\frac{223}{m}}

9.143=\sqrt{\frac{223}{m}}

Squaring both sides, we obtain the following;

9.143^2=\frac{223}{m}\\9.143^2*m=223\\83.592m=223\\therefore\\m=\frac{223}{83.592}\\m=2.67kg

8 0
3 years ago
Which group of words do you think is an idiom?
Lady bird [3.3K]

Explanation:

a chip on your shoulder is an example

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3 years ago
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