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3 years ago

What is the velocity of a car that went a distance of 400ft in 25 seconds

Physics

2 answers:

snow_tiger [21]3 years ago

5 0

Speed = (distance covered) / (time to cover the distance)

Speed = (400 ft) / (25 seconds)

<em>Speed = 16 ft/sec </em>

That's the car's <u><em>speed</em></u>.

We don't have enough information to state its <u><em>velocity</em></u>.

Rom4ik [11]3 years ago

3 0

Answer:

I believe it is 1/6 M

Explanation:

sorry if i am not right but i think i am

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Different structural forms of the same element are called

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Answer:

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3 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

amm1812

<u>Answer</u>:

The velocity of the tennis racket after the collision 14.966 m/s.

<u>Step-by-step explanation:</u>

let the following:

m₁ = mass of tennis racket = 0.311 kg

m₂ = mass of the ball = 0.057 kg

u₁ = velocity of tennis racket before collision = 30.3 m/s

u₂ = velocity of the ball before collision = -19.2 m/s

v₁ = velocity of tennis racket after collision

v₂ = velocity of the ball after collision

Right (+) , Left (-)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.

So, the total kinetic energy before collision = the total kinetic energy after collision.

So, 0.5 m₁ u₁² + 0.5 m₂ u₂² = 0.5 m₁ v₁² + 0.5 m₂ v₂² ⇒ (1)

Also, the total momentum before collision = the total momentum after collision.

So, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ ⇒ (2)

Solving (1) and (2):

∴ v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)

= ( 30.3 * (0.311 - 0.057) - 19.2 * 2 * 0.057 ) / ( 0.311 + 0.057)

= 14.966 m/s.

So, the velocity of the tennis racket after the collision 14.966 m/s.

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3 years ago

A piece of metal has a mass of 10g and a mass of 2cm

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Answer:

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Explanation:

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2 years ago

An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =

goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of reynold number for both objects

$R_{1}=R_{2}$

$\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}$

$\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}$

Here, $k_{1}=k_{2}$

$h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}$

$h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}$

Put the value into the formula

$h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}$

$h_{2}=20\ W/Km^2$

Hence, The value of the average convection coefficient is 20 W/Km².

7 0

3 years ago

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

1 year ago

What is the velocity of a car that went a distance of 400ft in 25 seconds ​

What is the velocity of a car that went a distance of 400ft in 25 seconds