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3 years ago

Find the velociity of a car which travels 35 m to the right over a period of 40 seconds

Physics

1 answer:

GalinKa [24]3 years ago

8 0

Answer:

the velocity of the car is 0.875 m/s

Explanation:

$v = d \div t \\ v = 35m \div 40sec = 0.875 \: m \: per \: sec$

therefore the V of car is 0.875 m

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(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation

Musya8 [376]

Answer:

the answee is

2.968456 ×10^7

4 0

3 years ago

The same pipe is used to carry both air and water. For the same fluid velocity and friction factor for the air and water flows:

VladimirAG [237]

Answer:

the pressure drop for the water flow is greater than that for the air flow.

Explanation:

Detailed analysis of the problem is show below.

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3 years ago

If the box is a distance 1.81 m from the rear of the truck when the truck starts, how much time elapses before the box falls off

alexira [117]

If the box is a distance 1.81 m from the rear of the truck when the truck starts, ... Force of Friction = mu_s * Normal Force( M * G) ... The box starts moving! ... Now that the box is moving, the bed of the truck pulls at it with 17.4 ... out how long it will take the box to reach the back of the truck. ... T^2 = 2 * 1.81 / .64

4 0

4 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago

4. The chart shows melting and boiling points of

kondor19780726 [428]

Answer:

The correct option is A

Explanation:

Firstly, it should be noted that the freezing point of a substance can be assumed to be melting point of that substance because a substance will normally change from liquid to solid (freezes) at the same point it changes from solid to liquid (melts). For example, water freezes at 0°C and also starts melting at 0°C.

Thus, the substance with the lowest melting point among the substances mentioned in the question is alcohol (ethanol) with the melting point of -114°C. Hence, ethanol also has the lowest freezing point thereby freezing at the lowest temperature.

6 0

3 years ago