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kkurt [141]
3 years ago
15

A block with mass m1 = 8.8 kg is on an incline with an angle θ = 41.0° with respect to the horizontal. For the first question th

ere is no friction, but for the rest of this problem the coefficients of friction are: μk = 0.39 and μs = 0.429.. . . . To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.13 m from its unstretched length. . . . . Now a new block with mass m2 = 16.4 kg is attached to the first block. The new block is made of a different material and has a greater coefficient of static friction is needed between the new block and the plane to keep the system from accelerating?
Physics
1 answer:
serg [7]3 years ago
6 0
So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2
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an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?
gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

F_{net} = F_g - F = m\cdot g - F\\

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N

The tension force F in the supporting cables is 7245N


3 0
4 years ago
A platinum ball weighing 100 g is removed from a furnace and dropped into 400 g of water at 0 degree C. If the equilibrium tempe
kakasveta [241]

Answer:

T = 1010 degree Celsius

Explanation:

mass of ball (Mb) = 100 g

mass of water (Mw) = 400 g

temp of water = 0 degree

specific heat of platinum (C) = 0.04 cal/g degree celsius

we can calculate the temperature of the furnace from the equation before

Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)

100 x 0.04 x ( T - 10) = 400 x (10 - 0)

4 (T - 10) = 4000

T - 10 = 1000

T = 1010 degree Celsius

3 0
4 years ago
1. You wish to heat 20 kg of water from 40°C to 80°C. How many kcal of heat are necessary to do this? To how many kJ does this c
adelina 88 [10]

Answer:

<h2>3,343.68kJ </h2>

Explanation:

Heat energy used up can be calculated using the formula:

H = mcΔt

m = mass oof the object (in kg) = 20kg

c = specific heat capacity of water = 4179.6J/kg°C

Δt change in temperature = 80-40 = 40°C

H= 20* 4179.6 * 40

H = 3,343,680Joules

H = 3,343.68kJ

8 0
3 years ago
This is physics class, plzzz help ! 86.72 m to cm
Vaselesa [24]

Answer:

8672

Explanation:

multiply the length value by 100

5 0
3 years ago
A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field
Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= -  .086 Weber

change in flux

.15 +  .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0
3 years ago
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