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Ede4ka [16]
3 years ago
5

Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

Physics
1 answer:
lana [24]3 years ago
8 0

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

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A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
zzz [600]

Answer:

0.0257259766982 m

Explanation:

P_2 = Atmospheric pressure = 101325 Pa

d_1 = Initial diameter = 1.5 cm

d_2 = Final diameter

\rho = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

The diameter of the bubble is 0.0257259766982 m

8 0
3 years ago
A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
Montano1993 [528]

Net Force = (mass) x (acceleration)  (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

6 0
2 years ago
If you weighed 130 pounds on earth, you would weigh _____pounds on the moon
Aneli [31]

Answer:

152 pounds

Explanation:

4 0
3 years ago
A worker pushed a 33 kg block 6.1 m along a level floor at constant speed with a force directed 23° below the horizontal. if the
jenyasd209 [6]
The work done occurs only in the direction the block was moved - horizontally. Work is given by:

W = F(h) * d

Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.

Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):

F(h) = F(app)cos(23)

Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:

N = mg + F(v) = mg + F(app)sin(23)

Now we can get down to business and solve for F(app) - as mentioned above:

F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8

Now that we have F(app), we can find the exact value of F(h):

F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7

And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3

Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
3 0
3 years ago
What is nuclear energy?
____ [38]

Answer: D. Energy released from atoms.

7 0
3 years ago
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