All categories

3 years ago

1. Force of impact is the force ________________.

Physics

1 answer:

siniylev [52]3 years ago

6 0

Force of impact is the force generated when objects meet.

You might be interested in

Which type of energy will definitely not be used in the lighting of a match?

In-s [12.5K]

I say that the answere would be B

4 0

3 years ago

When distances were carefully measured from the Earth to globular clusters above and below the Milky Way plane (where our view o

uranmaximum [27]

Answer:

A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction

Explanation:

The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation

4 0

3 years ago

(Will give brainliest)

Sveta_85 [38]

Answer:

b.) Length

Explanation:

The length of the string can be changed by removing it from the slotted bracket and placing it back in. You can change the mass by varying the number of washers on the mass hanger. The amplitude can be changed by varying the starting angle of the pendulum (low, medium, and high angle). sorry if wrong

8 0

3 years ago

Need help 14 points

valentinak56 [21]

Biological adaptation

3 0

3 years ago

Read 2 more answers

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human i

Varvara68 [4.7K]

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$ .

7 0

2 years ago