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Archy [21]
3 years ago
15

5. Find the mass of a car that is traveling at a velocity of 35 m/s West.

Physics
1 answer:
sattari [20]3 years ago
3 0

Answer:

m = 9795.9 kg

Explanation:

v = 35 m/s

KE = 6,000,000 J

Plug those values into the following equation:

KE = \frac{1}{2} mv^{2}

6,000,000 J = (1/2)(35^2)m

---> m = 9795.9 kg

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babymother [125]

Answer:

Option C

Explanation:

Given that

Motor force is 250 N

Force of friction is 750 N

Weight is 8500 N

And, the normal force is 8500 N

Now based on the above information

Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force

8 0
2 years ago
Which statement is true of the carbon atoms that make up a diamond?
Svetlanka [38]
B, this is because the particles in a solid such as the diamond can not move and even though they are locked into place they still vibrate
8 0
3 years ago
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How much energy to be removed to drop the temperature of 5.7kg of wood from 20degrees to 7degrees. #100points
Neporo4naja [7]

see i was trying to figure out the answer but i didn't understand it so i took the time to research and work it out but i still didn't understand i found one that was close to it and i got the same one as the other person which is D but idk if it is that type of question if it is than it is d if not then idk

5 0
3 years ago
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Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at
Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0
3 years ago
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

m_2 = \frac{20}{3} kg

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}

a = 0.981 m/s^2

so angular acceleration is given as

\alpha = \frac{a}{r_1}

\alpha = \frac{0.981}{0.5}

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = \frac{I\alpha}{r_1}

T = \frac{45 (1.96)}{0.5}

T = 176.6 N

7 0
3 years ago
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