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Maksim231197 [3]
3 years ago
10

Please help asap I need this rn

Physics
1 answer:
den301095 [7]3 years ago
6 0
It’s the third one .
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According to the general theory of relativity, what are consequences of the curvature of space-time? Check all that apply. The d
Svetlanka [38]

Answer:

The dilation of time.

The falling of objects.

The changing of paths of light.

Explanation:

I have explained in the image attached below.

From the explanation, the correct ones are;

The dilation of time.

The falling of objects.

The changing of paths of light.

4 0
3 years ago
Two people are pushing a couch from opposite sides. one person pushing to the right with a force of 3 N. the other person pushes
skelet666 [1.2K]

letter a because 4N to the left +3N to the right = 4N left minus 3N left as you change the direction of 3N then you also change the plus into minus. the direction will always favour to which of the direction is bigger

8 0
3 years ago
You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of
artcher [175]

Answer:

The distance is 1.026 m.

Explanation:

mass of rod, M = 1.23 kg

Length, L = 1.25 m

mass, m = 10 kg

Time period, T = 2 s

Let the distance is d.

The formula of the time period is given by

T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16  = 0.64 + 10d^2\\\\d= 1.026 m

3 0
3 years ago
An object is moving with an initial velocity of 19 m/s.It is then subject to a constant acceleration of 2.5 m/s for 15s.How far
bija089 [108]

Answer:

566.3 m

Explanation:

The distance travelled by the object can be found by using the SUVAT equation:

d=ut+\frac{1}{2}at^2

where

u is the initial velocity

t is the time

a is the acceleration

For the object in this problem:

u = 19 m/s

a = 2.5 m/s^2

Substituting t = 15 s, we find the distance travelled:

d=(19)(15)+\frac{1}{2}(2.5)(15)^2=566.3 m

7 0
4 years ago
You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
3 years ago
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