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2 years ago

2. Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The total force

Physics

1 answer:

Arte-miy333 [17]2 years ago

3 0

b) Equal to 243 N.

Explanation:

The total force acting on the car in the opposite direction including the road friction and air resistance is equal to 243 N.

This is in conformity with newton's third law of motion.

Newton's third law of motion states that "action and reaction are equal and opposite in direction. "

The action force is that of the pull by Harry acting on the car.
The reaction force is in the opposite direction.
Both action and reaction force equal and opposite and magnitude and direction

learn more:

Newton's laws brainly.com/question/11411375

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A 10-meter long ramp has a mechanical advantage of 5. What is the height of the ramp?

denpristay [2]

1. If the ramp has a length of 10 and has a mechanical advantage (MA) of 5. Then we need to find the height of the ramp.
Formula:
MA = L / H
Since we already have the mechanical advantage and length, this time we need to find the height .
MA 5 = 10 / h
h = 10 / 5
h = 2 meters

Therefore, the ramp has a length of 10 meters, a height of 2 meters with a mechanical advantage of 5.

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2 years ago

You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro

ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$

Substituting k=907.5 N/m and $\Delta x=8.80 cm=0.088 m$ , we find the amount of work that must be done:

$W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J$

5 0

2 years ago

A sleeping 68 kg man has a metabolic power of 79 w .

Lesechka [4]

65W * 8h * 3600s/h = 1.9e6 J = 447 Cal

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2 years ago

A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g an

leonid [27]

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

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2 years ago

Consider a U-tube whose arms are open to the atmosphere. Now water is poured into the U-tube from one arm, and light oil (r 5 79

EastWind [94]

Answer:

water height: 3cm

oil height: 12 cm

Explanation:

70 cm = 0.7 m

Suppose both arms have the same cross-section area S:

Let water density be $\rho_w = 1000 kg/m^3$

Let x be the height of water in the 2nd arm, then 4x is the heigh of oil in the 2nd arm.

For the system to be balanced, then the fluid mass of both arms must be the same

$m_1 = m_2$

$V_1\rho_w = V_w\rho_w + V_o\rho_o$

$Sh_1\rho_w = Sx\rho_w + S4x\rho_o$

Where V1 is the fluid volume is the 1st arm, Vw is the water volume in the 2nd arm, Vo is the oil volume in the 2nd arm, $h_1 = 0.7 m$ is the fluid height in the 1st arm, $\rho_o = 5790kg/m^3$ is the oil density.

We can divide both sides by S

$h_1\rho_w = x\rho_w + 4x\rho_o$

$h_1\rho_w = x(\rho_w + 4\rho_o)$

$x = \frac{ h_1\rho_w }{(\rho_w + 4\rho_o } = \frac{0.7*1000}{1000 + 4*5790} = 0.03m = 3 cm$

So the water height in the 2nd arm is 3 cm and the oil height is 3*4 = 12 cm

7 0

2 years ago