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Ierofanga [76]
2 years ago
10

A body floats at a volume of 2/5 when it is placed in oil, how dense the object is ....s=900km/m3​

Physics
1 answer:
Nookie1986 [14]2 years ago
6 0

Answer:

ohhh wowwwwwwwwwwwwwwwwwp

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you push a sled of mass 15 kg across the snow with a force of 180 N for a distance of 2.5 m. There is no friction. if the sled s
Oxana [17]
<span>7.7 m/s First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so 180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2 Now determine how long you pushed it. For constant acceleration the equation is d = 0.5 A T^2 Substitute the known values getting, 2.5 m = 0.5 12 m/s^2 T^2 2.5 m = 6 m/s^2 T^2 Solve for T 2.5 m = 6 m/s^2 T^2 0.41667 s^2 = T^2 0.645497224 s = T Now to get the velocity, multiply the time by the acceleration, giving 0.645497224 s * 12 m/s^2 = 7.745966692 m/s After rounding to 2 significant figures, you get 7.7 m/s</span>
5 0
3 years ago
A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist
galina1969 [7]

Answer:

0.04225 Nm

Explanation:

N = Force applied = 5 N

\mu = Coefficient of static friction = 0.65

d = Diameter of knob = 1.3 cm

r = Radius of knob = \frac{d}{2}=\frac{1.3}{2}=0.65\ cm

Force is given by

F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N

When we multiply force and radius we get torque

Torque on thumb

\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm

Torque on forefinger

\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm

The total torque is given by

\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm

The most torque that exerted on the knob is 0.04225 Nm

4 0
2 years ago
4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two
Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=mV_{o}+MU_{o} (2)

p_{f}=(m+M)V_{f} (3)

m=1200 kg is the mass of the first car

V_{o}=20 m/s is the velocity of the first car, to the North

M=1400 kg is the mass of the second car

U_{o}=-22 m/s is the mass of the second car, to the South

V_{f} is the final velocity of both cars after the collision

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg} (6)

Finally:

V_{f}=-2.61 m/s (7) This is the resulting velocity of the wreckage, to the south

7 0
3 years ago
How does using a ramp to load boxes into a truck make work easier?
larisa [96]
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8 0
3 years ago
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Use the following quote to answer the question.
White raven [17]

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Explanation: which is b

3 0
2 years ago
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