A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?

Answers

Answer 1

Answer:68.15m/s

Explanation:

Given:

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

Formula:

v₁²=v₁²+2a (x)

Set up:

=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]

Solution:68.15m/s


Related Questions

what is the meaning of the word physics​

Answers

Answer:

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

Explanation:

mark as brainliest

How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J

Answers

Answer:

B) 200 [J]

Explanation:

In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]

F = m*g

where:

m = mass = 5 [kg]

g = gravity acceleration = 10 [m/s^2]

F = 5*10 = 50 [N]

w = F*d

where:

F = force = 50 [N]

d = 4 [m]

w = 50*4 = 200 [J]

Please provide explanation!!!
Thank you.

Answers

Answer:

(a) 102 cm/s

(b) 0.490 cm²

Explanation:

(a) Use Bernoulli equation.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0

½ ρ v₁² + ρgh₁ = ½ ρ v₂²

½ v₁² + gh₁ = ½ v₂²

½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²

v = 102 cm/s

(b) The flow rate is constant.

v₁ A₁ = v₂ A₂

(25.0 cm/s) (2.00 cm²) = (102 cm/s) A

A = 0.490 cm²

Write a haiku
poem
explaining
why graphing
is useful.
If you are
able, share
your poem
with others.

Answers

Answer:

Explanation:

graphing is helpful

helps visualize the line

of your equation

One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.

Answers

Answer:

The value  is  [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

Explanation:

From the question we are told that  

   The speed of the airplane from the north is [tex]\frac{dN}{dt}  =  -181 \  km /hr[/tex]

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  [tex]\frac{dE}{dt}  =  -278 \  km/hr[/tex]

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  [tex]N  =  30 \  km[/tex]

   The distance of the westbound plane is  [tex]E =  15 \  km[/tex]

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    [tex]R^2  = N^2 + E^2[/tex]

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     [tex]2R\frac{dR}{dt} =  2N \frac{dN}{dt} +   2E\frac{dE}{dt}[/tex]

Here

     [tex]R = \sqrt{N^2 + E^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R =33.54 \ m [/tex]

    [tex]2(33.54) * \frac{dR}{dt} =  2( 30)*(-181)  +   2*15*(-278)[/tex]

=>   [tex] 67.08 * \frac{dR}{dt} =  -19200[/tex]

=>   [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

The rate of change of the distance between the planes is 286.23 km/hr.

The given parameters;

speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 km

The distance between the two planes is calculated by applying Pythagoras theorem as shown below;

[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]

The rate of change of the distance between the planes is calculated as follows;

[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]

Thus, the rate of change of the distance between the planes is 286.23 km/hr.

Learn more here:https://brainly.com/question/11488002

Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.

Answers

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

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