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Inessa05 [86]
1 year ago
6

A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1. 5 second

s. Calculate the force involved.
Physics
1 answer:
madreJ [45]1 year ago
8 0

The force involved is = 25200 Newton

Given the values in the questions,

Mass = 1400 kg

Speed or velocity of the car = 27 m/s

Time is given = 1.5 seconds

According to the formula of force,

⇒ Force = Mass x Acceleration

⇒ Force = 1400 x Acceleration ----- equation 1

Now to calculate the value of acceleration we will use,

⇒ Acceleration = (Velocity or Speed) / Time

⇒ Acceleration = 27 / 1.5

⇒ Acceleration = 18 m/s^{2}

Putting the value of acceleration in equation 1,

⇒ Force = 1400 x 18

⇒ Force = 25200 Newton

Therefore, the force involved is = 25200 Newton

To learn more about Force,

brainly.com/question/13191643

#SPJ4

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Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, \Delta x=0.03\ m

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

Initial energy = Final energy

\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2

Substituting all the values in above equation,

\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

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A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

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2 years ago
Which of the following is true about Viscosity of liquids:
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A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

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