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2 years ago

7.

Physics

1 answer:

Agata [3.3K]2 years ago

7 0

Answer:

7. the ball with a mass of 8.0kg

8.1225 kg

9.4.2 kg-m/sec

10.15m/sec

11.0.001kgm/sec

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The mass of a meteoroid was found to be 1.24 × 107 kilograms. Write the mass in standard notation.

Musya8 [376]

Answer:

The value we are given in the question is 1.24 * 10^7. This form of writing number is called scientific notation. The standard notation is the normal, regular way of writing numbers. Scientific notation and standard notations are interchangeable.

1.24 * 10^7 written in standard notation will be = 1.24 * 10000000 = 12400000.

Thus, the mass of the meteoroid was 12400000 kg.

Explanation:

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3 years ago

In which scenario will the two objects have the greatest gravitational force

Brilliant_brown [7]

Answer: I think the answer C

Explanation:

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2 years ago

Read 2 more answers

If we have less power, we most likely have

boyakko [2]

The same amount of work being done over a long period of time!

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3 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity $I$ = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity $I$ is proportional to 1/(distance)²

i.e

$I$ ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e $I$ ₂ = $I$ ₁/2

Hence,

$I$ ₂/ $I$ ₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0

3 years ago