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Dmitrij [34]
3 years ago
10

2. Which activity requires a scientist to use

Physics
2 answers:
Alecsey [184]3 years ago
6 0

Answer:

c please mark me as brainiest

Explanation:

harina [27]3 years ago
5 0

Answer:

<h3>Sorry i didn't know about this question and sorry please don't mind...</h3>

Explanation:

<h3> bro can you please help me can you please mark's me as brainliests please...</h3>

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A football field is about 100m long. If it takes a person 20s to run its length, how fast were they running?
Olenka [21]
100 m/ 20 s= 5 m/s.

They were running at a speed of 5 m/s~
7 0
3 years ago
Calculate the total mechanical energy of a 2 kg duck flying at 5 m/s, at a height of 10 meters above the ground.
dolphi86 [110]

Answer:

Total mechanical energy = 225 J

Explanation:

Given:

Mass of duck (m) = 2 kg

Speed of duck (v)= 5 m/s

Height of duck from ground (h) = 10 m

Gravitation acceleration (g) = 10 m/s²

Find:

Total mechanical energy

Computation:

Total mechanical energy = Kinetic energy + Potential energy

Total mechanical energy = (1/2)mv² + mgh

Total mechanical energy = (1/2)(2)(5)² + (2)(10)(10)

Total mechanical energy = 25 + 200

Total mechanical energy = 225 J

5 0
3 years ago
A 1.3-kg ball is attached to the end of a 0.8-m string to form a pendulum. This pendulum is released from rest with the string h
Natali5045456 [20]

Answer:

2.9 m/s

Explanation:

Momentum will be conserved

Speed of the ball just before collision is

v = √2gh = √(2(9.8)(0.8)) = 3.96 m/s

The initial momentum is 1.3(3.96) = 5.15 kg•m/s

The block takes away momentum of 0.6(2.2) = 1.32 kg•m/s

Leaving the ball with momentum of 5.15 - 1.32 = 3.83 kg•m/s

vf(ball) = 3.83 / 1.3 = 2.946... ≈ 2.9 m/s

3 0
3 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
3 years ago
Please can someone answer this question ​
EastWind [94]
B????????? I thinkkkkk
7 0
3 years ago
Read 2 more answers
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