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3 years ago

7. A 1000 kg car is rolling down the street at 2.5 m/s. How fast would a 2500 kg car have to

Physics

1 answer:

babunello [35]3 years ago

7 0

1 m/s

Explanation:

To solve this question we use the following formula:

momentum = mass × velocity

momentum of the first car = 1000 kg × 2.5 m/s

momentum of the second car = 2500 kg × X m/s

To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.

momentum of the first car = momentum of the second car

1000 kg × 25 m/s = 2500 kg × X m/s

X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s

Learn more about:

momentum

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You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun us

ZanzabumX [31]

Answer:

The answer is

A. Pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

Explanation:

The question is incomplete, here is a complete question with full options

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun uses a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because:_________.

A. pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

B. viscous drag between the walls of the tip and the caulk causes the caulk to swirl around chaotically.

C. Newton’s third law requires most of the energy in the caulk to be used to push back on the plunger rather than moving it through the tip.

D. the high density of the caulk impedes its flow through the small opening.

Since the caulk is thick and the exit nozzle is small, the pressure needed to deliver the caulk will be very high as pressure is uniformly distributed at the plunger side at every part of the caulk, hence very high pressure is needed to deliver the caulk which is why the handle needed the very hard squeeze

3 0

3 years ago

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (

Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

$x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}$

$10 + 8.3 x = 0$

$x = -1.20 m$

Similarly for y direction we have

$y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}$

$9.28 + 8.3 x = 0$

$x = -1.12 m$

5 0

3 years ago

Can someone please help meee .

fiasKO [112]

Answer:

32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice

6 0

3 years ago

Read 2 more answers

A student who was training for a cross country race jogged for 2. 0 hours and covered a distance of 14. 0 kilometers. What was t

FinnZ [79.3K]

The average speed is 116.66m/s.

Given - The path traced is 14km , time for jogging is 2 hrs=120min

To find the average speed-

Speed refers to the ease of the movement and degree of mobility as a result of force application.
Due to this there is involvement of velocity.
Journey of average speed is the cumulative of distances and time.
Kinetic theory refers to the Boltzmann constant connecting to the standards of distance traversed.

calculations-

$Speed= \frac{Distance}{Time}$

Speed= 14 000 / 120

= 116.66m/s

To learn more about average speed -

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6 0

1 year ago

A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb

marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

$n_1 =\dfrac{46 \times 1}{0.0821\times 298}$

n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

$n_2 =\dfrac{12 \times 1}{0.0821\times 298}$

n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

$P_1=\dfrac{n_1 R T}{V}$

$P_1=\dfrac{1.89\times 0.0821\times 298}{5}$

P₁ = 9.25 atm

Partial pressure of oxygen

$P_2=\dfrac{n_2 R T}{V}$

$P_2=\dfrac{0.49\times 0.0821\times 298}{5}$

P₂ = 2.39 atm

total pressure

P = P₁ + P₂

P = 9.25 + 2.39

P = 11.64 atm

8 0

3 years ago