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3 years ago

What happens to the force needed to stretch an elastic object?

Physics

1 answer:

avanturin [10]3 years ago

8 0

Answer:

the force pulls each other cause its just physics

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A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with

Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 + 20 sin (30) t - 4.9 t^2

when it hits ground this = 0

0 = -4.9t^2 + 20 sin (30 ) t + 45

0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec

max height is at t= - b/2a = 10/9.8 =1.02

use this value of 't' in the equation to calculate max height = 50.1 m

it has 4.22 - 1.02 to free fall = 3.2 seconds free fall

v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s

total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m

8 0

2 years ago

What is the formula for silver nitrate

serious [3.7K]

Answer:

Explanation:

Agno3

8 0

3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

5 0

3 years ago

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

wlad13 [49]

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

5 0

3 years ago

Find the mass by adding the numbers from the three sliders

shtirl [24]

add the numbers from the three sliders to determine that mass of an object

3 0

3 years ago