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Nutka1998 [239]
3 years ago
15

If a mass on a spring is 16 kg and the spring constant is 4 N/m, what would be its period?

Physics
1 answer:
MArishka [77]3 years ago
8 0

Explanation:

T = 2π√(m/k)

T = 2π√(16 kg / 4 N/m)

T = 4π s

T ≈ 12.6 s

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A generator converts mechanical energy into _____________________ energy.
erica [24]
Electrical energy is your answer.
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3 years ago
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Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
3 years ago
You use a lever to lift a heavy tree branch you apply a force of 50 n and the lever lifts the branch
valentinak56 [21]

1.8 is the mechanical advantage of the lever.

<h3>Definition of mechanical advantage</h3>

The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.

The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.

Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.

In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.

Mechanical advantage= 90/50 =1.8

Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?

To learn more about the Mechanical advantage visit the link

brainly.com/question/16617083

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5 0
2 years ago
a grinding wheel is initially rotating with an angular velocity 5500 rad/srad/s when its motor is suddenly turned off. it comes
kiruha [24]

The angle through which the grinding wheel rotates in the first second =  <u>5300 rad</u>

Angular velocity is, the time charge at which an object rotates, or revolves, about an axis, or at which the angular displacement between our bodies changes. within the discern, this displacement is represented via the angle θ among a line on one body and a line on the alternative.

The angular velocity is described as the charge of trade of the angular position of a rotating body. Linear speed is defined because the charge of change of displacement with respect to time whilst the item moves alongside a straight course.

Initial angular velocity of the grinding wheel = ω1 = 5500 rad/s

Final angular velocity of the grinding wheel = ω2 = 0 rad/s   (Comes to rest)

Time is taken by the grinding wheel to come to rest = T = 10 sec

Angular acceleration of the grinding wheel = α

2 = ω1 + αT

0 = 5500 + α(10)

α = - 400 rad/s2

Negative as it is deceleration.

The angle through which the grinding wheel rotates in the first second = θ

Time period = T1 = 1 sec

θ = ω₁T1 + αT1²/2

θ = (5500)(1) + (-400)(1)²/2

θ = 5300 rad

The angle through which the grinding wheel rotates in the first second =  <u>5300 rad</u>

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5 0
1 year ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
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