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ale4655 [162]
4 years ago
13

The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.

Physics
1 answer:
Cerrena [4.2K]4 years ago
4 0

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g                  ( g=9.81 m/s² )

F=ma

  = (1510)*( 0.75*9.81)

  = 11109.825 N

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A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
4 years ago
A force of 10N acts on a car for 3 seconds. The impulse imparted on the car is
ss7ja [257]

Answer:

30N*s

Explanation:

Given the following data;

Force = 10N

Time = 3 seconds

To find the impulse;

Impulse = force * time

Substituting into the equation, we have;

Impulse = 10 * 3

Impulse = 30Ns

7 0
3 years ago
A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

Hence, it is the required solution.

7 0
3 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
Dmitrij [34]

Answer:

\lambda_2 = 573.3 nm

Explanation:

As we know that the position of maximum intensity on the screen is given as

y = \frac{N\lambda L}{d}

here we know that

\lambda = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}

so here we have

8 (645 nm) = 9 \lambda_2

\lambda_2 = 573.3 nm

4 0
3 years ago
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
8 0
3 years ago
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