All categories

3 years ago

A body of mass 500kg moving at a speed of 10m/s reaches the speed of 50m/s in 20s.The force exerted is

Physics

1 answer:

Leona [35]3 years ago

6 0

Answer:

answer is 1000 N

formula used-

F= m x (v-u/t)

mark this as brainliest!

You might be interested in

Explain two reasons why astronomers are continually building larger and larger telescopes. Explain two reasons why astronomers a

Sergio [31]

Answer:

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* arger mirrors decreases the angle of dispersion giving a better resolution of the bodies

Explanation:

Refracting telescopes get bigger every day for two main reasons.

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* the diffraction process for circular apertures is given by

θ = 1.22 λ / D

where d is the diameter of the mirror, therefore having larger mirrors decreases the angle of dispersion giving a better resolution of the bodies

5 0

3 years ago

Iron ball sinks in water but not in mercury. Why?

Zigmanuir [339]

Answer:

mercury is more dense

Explanation:

8 0

3 years ago

Read 2 more answers

1–Find the components of each vector

navik [9.2K]

Answer:

ANSWERS ARE 100 % correct

Explanation:

trust me

7 0

2 years ago

Which of the following is NOT common of elite Shang burials?

Zigmanuir [339]

Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

6 0

2 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

2 years ago