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EleoNora [17]
1 year ago
7

A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from it

s initial position. The rock is pulled back with a force of 10 newtons.
When the rock is released, what is its kinetic energy?
Physics
1 answer:
irakobra [83]1 year ago
4 0

Answer:

id

Explanation:

i don't know

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An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48
kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

Where:

Er = remaining energy [J]

Er = 108 [J]

3 0
2 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
Read 2 more answers
flat block is pulled along a horizontal flat surface by a horizontal rope perpendicular to one of the sides. The block measures
Lelu [443]

Answer:

Answered

Explanation:

v= 1 m/s

A= 1 m^2

m= 100 kg

y= 1 mm

μ = ?

ζ= viscosity of  SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2

forces acting on the block are  

                                 F_s   ←    ↓     →F_f

                                                mg

N= mg

F_s=  shear force = ζAv/y        F_f= friction force = μN

now in x- direction F_s= F_f

ζAv/y  =  μN

0.3075×1×1×1/1×10^{-3} = μ×100

⇒μ=0.313 (coefficient of sliding friction for the block)

Now, as the velocity is increased shear force also increases and due to this frictional force also increases.

Now, to compensate this frictional force friction coefficient must increase

as v∝μ

7 0
2 years ago
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A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistance
Nookie1986 [14]
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m
4 0
2 years ago
If the resistivity of copper is less than that of gold at room temperature, which of the following statements must be true? Gold
KiRa [710]

Answer:

Gold Has A Higher Resistance Than Copper. The Sample Of Gold Is Thinner Than The Sample Of Copper. Electrons In Gold Are More Likely To Be Scattered Than Electrons In Copper At Room Temperature When they are exelerated by the same electric field.

Explanation:

5 0
2 years ago
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