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ki77a [65]
2 years ago
5

To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torqu

e exerted by the weight about his knee, 40 cm away from the weight? Use g = 10 m/s2. A. 160 Nâm B. 16 Nâm C. 1600 Nâm D. 1.6 Nâm
Physics
1 answer:
Masja [62]2 years ago
8 0

Answer:

B. τ = 16 Nm

Explanation:

In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:

τ = Fd

here,

τ = Torque = ?

F = Force exerted by the weight = Weight = mg

F = mg = (4 kg)(10 m/s²) = 40 N

d = distance from knee to weight = 40 cm = 0.4 m

Therefore,

τ = (40 N)(0.4 m)

<u>B. τ = 16 Nm</u>

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The drawing shows a side view of a swimming pool. The pressure at the surface of the water is atmospheric pressure. The pressure
Nikitich [7]

Answer:

A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Explanation:

Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is

    P =P_{atm} + ρ g y

Where P_{atm} is the atmospheric pressure, ρ  the water density, and 'y' the depth measured from the surface.

Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth

Now we can examine the claims

A) To true. State agreement or with the equation above

B) False. Pressure changes with atmospheric pressure

C) False. It's the opposite

D) False. They are directly proportional

7 0
2 years ago
Read 2 more answers
At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m
Margaret [11]

Answer:

Part a)

t = \sqrt{\frac{2h}{g}}

Part b)

t = 1.06 s

Part c)

L  = 4.86 m

Explanation:

Part a)

The height of the diving board is given as

h = 5.5 m

now the speed of the diver is given as

v_0 = 2.7 m/s

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

y = v_y t + \frac{1}{2}at^2

h = 0 + \frac{1}{2}gt^2

t = \sqrt{\frac{2h}{g}}

Part b)

t = \sqrt{\frac{2h}{g}}

plug in the values in the above equation

t = \sqrt{\frac{2(5.5 m)}{9.81}

t = 1.06 s

Part c)

Horizontal distance moved by the diver is given as

d = v_0 t

d = 2.7 \times 1.06

d = 2.86 m

so the distance from the edge of the pool is given as

L = 2.86 + 2

L  = 4.86 m

4 0
2 years ago
A submarine completed a 450 km training with an average speed of 50 km/h. For the first 180 km, it travelled at an average speed
Kryger [21]

Answer:

45km/hr

Explanation:

Total distance=450km

Total speed=50km/hr

Total time= distance/speed

=450/50

=9hrs

distance a=180km

speed a=60km/hr

Time a=180/60

=3hrs

Distance b=450-180=270km

Speed b=?

Time b=270/speed b

Total time=time a + time b

9=3+(270/speed b)

270/speed b =9-3

270/speed b =6

6*speed b =270

Speed b=270/6

Speed b=45km/hr

4 0
2 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
ycow [4]
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius  is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
  = 0.5*(0.15 kg)*(0.038 m)²
  = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
       = 0.5*(0.15 kg)*(0.5 m/s)²
       = 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
       = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
       = 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J


4 0
2 years ago
Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is
finlep [7]

Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

The speed of the cart after 3 seconds of Medium fan speed is   36.0 cm/s

The speed of the cart after 6 seconds of High fan speed is  96.0 cm/s

Explanation:

took the test on edgenuity

4 0
2 years ago
Read 2 more answers
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