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Natasha_Volkova [10]
3 years ago
9

What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied

Physics
2 answers:
Greeley [361]3 years ago
6 0

Answer: The spring constant for the supermarket scale is 400 N/m.

Explanation:

To calculate the spring constant, we use Hooke's Law. The formula for Hooke's law is:

F=kx

where, F = Force applied on the spring = 4N

k = Spring constant = ? N/m

x = Compression or extension of the spring = 0.01m

Putting values in above equation, we get:

400N=k\times 0.01m\\\\k=400N/m

Hence, the spring constant for the supermarket scale is 400 N/m.

Nikolay [14]3 years ago
3 0
The answer to the question is 400 N. 
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Una barra de aluminio que esta a 78 GRADOS CENTIGRADOS entra en contacto con una barra de cobre de la misma longitud y área que
stiks02 [169]

Answer:

Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

Explanation:

Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:

(78 + 28) / 2 = X

106 / 2 = X

53 = X

Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

8 0
3 years ago
What makes music different from noise?
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5 0
3 years ago
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Describe in terms of kinetic and potential energy what happens if an apple falls from a tree and comes to rest on the ground( wr
Lunna [17]

Answer:

An apple hanging at a branch has potential energy due its position. It can be written as PE= mgh where m is the mass of the apple h is the distance between the apple and the ground and g is the acceleration due to gravity.

as the apple falls from the tree it loses its potential energy and gains kinetic energy due to the movement of the apple. Its kinetic energy will be given by KE= 1/2mv²  where m is the mass of the apple and v is the speed with which the apple falls.

As the apple falls the height or the distance reduces and PE becomes reduces. But it gains Kinetic energy due to its speed.

But when the apple falls to the ground and comes to rest its kinetic energy is converted to potential energy.

thus the total energy remains the same. it changes from one form to the other but remains unaltered.

6 0
3 years ago
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
4 years ago
A car has an acceleration of 1.5 m/s2. A net force of 2100 N is acting on the car. What is the mass of the car?
gulaghasi [49]

Answer:

Given

acceleration (a) =1.5ms2

Force(F) =2100N

R. t. c mass (m) =?

Form

F=ma(divided by m both sides)

m=F/a

m=2100/105

m=1400kg

mass of car =1400kg

8 0
3 years ago
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