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vova2212 [387]
3 years ago
13

a 2.5 kg rock is dropped off a 32 m cliff and hits a spring, compressing it 57cm. what is the spring constant

Physics
2 answers:
Rama09 [41]3 years ago
4 0

2.5 kg because you cant change the weight of the rock

Mila [183]3 years ago
3 0

The answer is 4,800N/m

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You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc
pickupchik [31]

a) 15.4^{\circ}

b) 5.2 m/s

c) 151.2 N

Explanation:

a)

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

mg sin \theta

where

m =16 kg is the mass of the box

g=9.8 m/s^2 is the acceleration due to gravity

\theta is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

F=ma\\mgsin \theta = ma

where a is the acceleration.

From the equation above we get

a=g sin \theta

And we are told that the acceleration must not exceed

a=2.6 m/s^2

Substituting this value and solving for \theta, we find the maximum angle of the ramp:

\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}

b)

Here we are told that the vertical distance of the ramp is

h=1.4 m

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

GPE=KE\\mgh=\frac{1}{2}mv^2

where:

m = 16 kg is the mass of the box

g=9.8 m/s^2

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s

c)

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

mg cos \theta

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

N-mg cos \theta

and by substituting:

m = 16 kg

g=9.8 m/s^2

\theta=15.4^{\circ}

We can find the normal force:

N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N

8 0
3 years ago
Will someone help me please :(
-Dominant- [34]
The answer's 25,000 joules 
half m v squared
half 2000, x 5 squared
7 0
3 years ago
Which one of the following substances is a liquid fuel used in rocket engines?
sergiy2304 [10]
The answer is, "B", "Ammonia".
4 0
3 years ago
Read 2 more answers
A man pulls a 50.0 kg box with a rope parallel to the ground he pulls the box 10.0 m about how much work has he done
anygoal [31]
Your answer is 5000 J

when W(work) = F X  when F= the force and X= the displacment

and F(g) = M a(g) when M= mass and a = the acceleration and in our question
, the force is the gravitational force and a= 9.8 m/S2 we can assume as 10 m/s2

and when we have M= 50 Kg
so by substitution:
F= 50 x 10 = 500 N

and by substitution in work equation: when x = 10 m 
∴ W = 500 x 10 = 5000 j

4 0
3 years ago
Read 2 more answers
When the body requires an increased blood flow rate in a particular organ or muscle, it can accomplish this by increasing the di
horsena [70]

Answer: The percentage increase in diameter is 19%

Explanation:

<em>Using Poiseuille's law which states that Flow rate , Q = ΔPπr⁴/8ηl</em>

<em>where ΔP is pressure difference, π is a constant, r is the radius which is half of the diameter, η is viscosity of blood, l is length of blood vessel</em>

Let Q₁ be the intitial flow rate, Q₂ the final flow rate,r₁ and r₂ the initial and final radius.

Note: r = d/2, therefore, r₁⁴ = d₁⁴/16 r₂⁴ = d₂⁴/16

Also, Q₂ = 2Q₁, since the flow rate is doubled

Since all factors except diameter is constant, d₂⁴/16 = 2d₁⁴/16

d₂⁴/16 =  d₁⁴/8

multiply both sides of the equation by 16

d₂⁴ = 2d₁⁴

taking the fourth root of both sides

d₂ = 1.19*d₁

d₂ - d₁ = 1.19d₁ - d₁

d₂ - d₁ = d₁(1.19 - 1)

d₂ - d₁ = 0.19d₁

d₂ - d₁/d₁ = (0.19d₁/d₁)*100%

(d₂ - d₁/d₁)*100% = 19%

Therefore, the percentage increase in diameter is 19%

3 0
4 years ago
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