What determines the life cycle of a star?

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

IRINA_888 [86]

Answer:

s=4.44 m

Explanation:

Given that

Coefficient of the kinetic friction ,μ = 0.13

Initial velocity ,u= 3.4 m/s

Final velocity of the box ,v= 0 m/s

The acceleration due to friction force

a= - μ g

Now by putting the values in the above equation

a= - 0.13 x 10 ( take g= 10 m/s²)

a= - 1.3 m/s²

We know that

v²= u ² + 2 a s

s=distance

a=acceleration

v=final speed

u=initial speed

Now by putting the values in the above equation

0²= 3.4² - 2 x 1.3 x s

$s=\dfrac{3.4^2}{2\times 1.3}\ m$

s=4.44 m

The distance cover by box will be 4.44 m.

6 0

3 years ago

Pls what is a friendly relationship

Anuta_ua [19.1K]

Answer:

A relationship where both members are freindly and supportive towards each other.

Explanation:

Brainliest?

4 0

4 years ago

Read 2 more answers

When a potential difference of 12 v is applied to a wire 6.8 m long and 0.35 mm in diameter the result is an electric current of

Nonamiya [84]

The total resistance is R = voltage / current

This resistance R = (L/A)r where:
r is the resistivity
L is the length of the conductor
A is the cross-sectional area of the conductor.

thus r = RA/L

You are given L; you can compute R from the voltage and current you are given; and the cross-sectional area of a round wire is (pi)(radius^2) or (pi/4)(diameter^2)

8 0

3 years ago

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

5 0

4 years ago

When the input work on simple machine equals the work the machine is said to be 100% efficient

Semmy [17]

Answer:

This is not a question, but a statement. The concept that the work done is equal to the input work on a simple machine is called ideal machine. An ideal machine has 100% efficiency which means that there is no loss. All the amount of input work is equal to the useful output of the machine.

Explanation:

8 0

3 years ago