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3 years ago

What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?

1 answer:

4 0

Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)

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How does`ezzzzz321erfq23c2f

Nataliya [291]

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7 0

2 years ago

The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

Sindrei [870]

Answer with Explanation:

We are given that

Speed , $v_0$ =107257 kmph

Angular velocity, $\omega=7.292\times 10^{-5} rad/s$

Radius of earth,r= $6371 km=6371000 m$

1 km=1000 m

Linear velocity, $v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s$

Linear velocity, $v=464.57\times \frac{18}{5}=1672.46 km/h$

Velocity at point A, $v_A=-vi+v_0 j=-1672.46 i+107257j kmph$

Velocity at point B, $v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph$

Velocity at point C, $v_C=v_0j+vi=1672.46 i+107257j kmph$

Velocity at point D, $v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph$

3 0

3 years ago

Not sure if it went through last time. Please help asap!

Olin [163]

The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get
$0.42 \times 14.8 = 6.22$
The answer is 6.22 N because newtons are the unit used to measure force.

8 0

3 years ago

Read 2 more answers

A centrifuge is a device used to separate materials by their masses. A sample in a centrifuge is rotated at high speeds along a

german

Answer:

F = 1.047 10⁻² N

Explanation:

Let's use kinematics to find the angular acceleration

w = w₀ + α t

as for rest w₀ = 0

w = α t

α = w / t

let's reduce the magnitudes to the SI system

w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s

m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg

r = 10.0 cm (1 m / 100 cm) = 0.100 m

let's calculate

α = 104.72 / 1

α = 104.72 rad / s²

angular and linear variables are related

a = α r

a = 104.72 0.100

a = 10.47 m / s²

finally we substitute in Newton's second law

F = 1 10⁻³ 10.47

F = 1.047 10⁻² N

8 0

3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$