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4 years ago

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An object has a mass of 20 g and a volume of 5 cm3. what is the object's density? 4 g/cm3 100 g/cm3 0.25 g/cm3

1 answer:

Sati [7]4 years ago

8 0

The answer is 110 g/cm3

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If a car travels 216 kilometers in 4 hours, calculate its speed.

Mariana [72]

Answer:

15 m/s

Explanation:

Speed(m/s) = distance(m)/time(s)

distance = 216 km = 216,000 m

time = 4 hours = 14,400 s

speed = 216000/14400 = 15 m/s

6 0

2 years ago

Read 2 more answers

ivanzaharov [21]

Answer:

1 +2=3 so thats it

Explanation:

Because i need points

4 0

3 years ago

Why does a satellite in a circular orbit travel at a constant speed? why does a satellite in a circular orbit travel at a consta

motikmotik

Why does a satellite in a circular orbit travel at a constant speed? why does a satellite in a circular orbit travel at a constant speed? there is a force acting opposite to the direction of the motion of the satellite. there is no component of force acting along the direction of motion of the satellite. the net force acting on the satellite is zero. the gravitational force acting on the satellite is balanced by the centrifugal force acting on the satellite?
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3 years ago

Proposed Exercise - Circular Movement

notka56 [123]

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

5 0

3 years ago

A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm

qaws [65]

Answer:

0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

0.05925 N

$29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

Explanation:

I = Current

B = Magnetic field

Separation between end points is

$l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm$

Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

The direction is $29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

4 0

3 years ago