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Vaselesa [24]
3 years ago
14

A car has a mass of 1.20 x 10^3 kilograms and a momentum of 2.00 x 10^4 kilogram meters/second. What is the velocity of the car?

Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0
The answer will be v=16.7m/s
I hope this helps you, have a great day!
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A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>​
Vadim26 [7]

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

7 0
2 years ago
Good morning friends​
Hatshy [7]

Answer:

Good morning. Whats about you

4 0
3 years ago
Read 2 more answers
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

4 0
3 years ago
according to this passage when making healthcare decisions, how should a person approach the decision making process
Aleksandr-060686 [28]

Answer:

What is your name?

Explanation:

no bro I don't know I am New of this app so I don't know

8 0
2 years ago
A car of mass 600 Kg is moving at 15m/s. Calculate its momentum.
mr_godi [17]
Mass=600kg
Velocity =15m/s
Momentum(p)=?
Now,
P=mass x velocity
=600x15
=9000kgm/s
3 0
3 years ago
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