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3 years ago

12

Electric field lines show the strength and _____ of an electric field.

1 answer:

Gwar [14]3 years ago

8 0

So we want to know what do electric field lines show except strength of the electric field. Electric field lines are lines that show us the field strength and the DIRECTION of the electric field. If the lines point away from the charge that is producing the lines, the charge is positive, and if the lines point towards the charge that is producing them that charge is negative.

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Newton's Principle states that the buoyant force on an object is equal to weight of the fluid displaced by the object.

Ilia_Sergeevich [38]

Answer:

False it wasn't Newton who theorised it, it was Archimedes

Explanation:

3 0

2 years ago

6. Which does more wor:

umka2103 [35]

Explanation:

1N=9,81 kg

1. 9,81×500×10=<u>49.050</u>

2. 9,81×100×40=39.240

option 1 does more wor

8 0

2 years ago

A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.

Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' = 19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0

3 years ago

Physics Question 100 Points No Spam.

VARVARA [1.3K]

Answer:

- Wind resistance made decrease in speed

-Gravity/Mass made decrease in velocity

Explanation:

7 0

2 years ago

Read 2 more answers

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

2 years ago