All categories

3 years ago

ASAP 1-5 question plzz helppp, mark Brainliest if you answer all 5 questions

Physics

2 answers:

podryga [215]3 years ago

7 0

Answer:

a,b,d,b, false

Explanation:

Licemer1 [7]3 years ago

4 0

1)a
2)b
3)d
4)b
5) false

You might be interested in

What is the gravitational force between two 1 kg objects that are 1 m apart?

lapo4ka [179]

Answer:

$F=6.67\times 10^{-11}\ N$

Explanation:

Given that,

The masses of two objects, m₁ = m₂ = 1 kg

The distance between masses, d = 1 m

We need to find the gravitational force between two masses. The force is given by the relation as follows :

$F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{1\times 1}{(1)^2}\\F=6.67\times 10^{-11}\ N$

So, the force between two masses of 1 kg is $6.67\times 10^{-11}\ N$ .

3 0

3 years ago

How does friction affect energy

Lera25 [3.4K]

Answer:

Friction is a resistive force to motion. When two bodies move against each other some of the kinetic energy is converted to heat energy due to friction. This reduces the total kinetic energy in the system.

Explanation:

5 0

3 years ago

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

3 0

3 years ago

Which statement is part of the principle of conservation of charge?

fgiga [73]

It is D. An object can acquire a net charge only when charges are transferred to or from it.

4 0

3 years ago

Read 2 more answers

A camera phone called the iPhone X has an image sensor size of 5 mm and a focal length of 4 mm. How long of a selfie-stick must

azamat

Answer:

length of selfie-stick is 1.62 m

Explanation:

Given data

image size h1 = 5 mm = 5 × $10^{-3}$ m

focal length = 4 mm = 4 × $10^{-3}$ m

distance h2 = 2.032 m

to find out

How long of a selfie-stick

solution

here we find first magnification

that is M = h1 /h2

M = 5 × $10^{-3}$ / 2.032

M = 2.46 × $10^{-3}$

and we know M = p/q

so p = Mq = 2.46 × $10^{-3}$ q

so we apply lens formula

1/f = 1/p - 1/q

1/ 4 × $10^{-3}$ = 1 / 2.46 × $10^{-3}$ q - 1/q

q = 1.622 m

so length of selfie-stick is 1.62 m

4 0

3 years ago