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Zina [86]
3 years ago
9

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

2.4 N. The free-body diagram shows the forces acting on the sled.
What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?

Physics
1 answer:
ratelena [41]3 years ago
8 0

Answer:

a = 1.3 m/s2; fn = 63.1 n :)

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A mass hanging from a spring is set in motion and its ensuing velocity is given by v (t )equals 2 pi cosine pi t for tgreater th
lianna [129]

Answer:

2(maximum), -2(minimum), -2(maximum).

Explanation:

V(t)= 2πcos πt--------------------------------------------------------------------------------(1).

Therefore, there is a need to integrate v(t) to get S(t).

S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).

Applying the condition given, we have s(0)= 0.

S(0)= 2sin ×π(0) + C.

Which means that; 0+C= 0. That is; C=0.

S(t)= 2 sin πt.

The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.

Take note that;  sin(π/2) = sin(5π/2) = 1 .

Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.

Therefore, we have that 2 maximum; -2 minimum and -2 maximum.

7 0
3 years ago
How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

___________________________________________

coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

8 0
3 years ago
A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find
FrozenT [24]

Answer:

The time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

Explanation:

Given :

Current I = 0.106 A

Area of plate A = 36 \times 10^{-4} m^{2}

Plate separation d = 4 \times 10^{-3} m

(A)

First find the capacitance of capacitor,

   C = \frac{\epsilon _{o} A }{d}

Where \epsilon _{o} = 8.85 \times 10^{-12}

   C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4}  }{4 \times 10^{-3} }

   C = 7.9 \times 10^{-12} F

But   C = \frac{Q}{V}

Where Q = It

  C = \frac{It}{V}

  V = \frac{It}{C}

Now differentiate above equation wrt. time,

  \frac{dV}{dt} = \frac{I}{C}

       = \frac{0.106}{7.9 \times 10^{-12} }

       = 1.34 \times 10^{10} \frac{V}{s}

Therefore, the time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

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