Ask question

Ask question

All categories

2 years ago

15

a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed

of the car is

1 answer:

Juliette [100K]2 years ago

8 0

Mass of the car (m) = 2000 Kg
Initial velocity (u) = 15 m/s
Force (F) = 10000 N
Time (t) = 3 s
Let the acceleration be a.
By using the formula, F = ma, we get,
10000 N = 2000 Kg × a
or, a = 10000 N ÷ 2000 Kg
or, a = 5 m/s^2
Let the final velocity be v.
By using the formula, v = u + at, we get,
v = 15 m/s + 5 m/s^2 × 3 s
or, v = 15 m/s + 15 m/s
or, v = 30 m/s

<u>Answer</u><u>:</u>

<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

You might be interested in

A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?

ivolga24 [154]

Answer:

Explanation:

Given

mass of rock $m=40\ kg$

Elevation of Rock $h=10\ m$

Distance traveled by rock with time

$h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

$5=0\times t+\frac{1}{2}(9.8)(t^2)$

$t^2=\frac{10}{9.8}$

$t=1.004\approx 1\ s$

velocity at this time

$v=u+at$

$v=0+9.8\times 1.004$

$v=9.83\ m/s$

6 0

3 years ago

Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

alina1380 [7]

Answer:

Datos:

q1 = -50 μC = $50*10^{-6}$

q2 = +30 μC = $30*10^{-6}$

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = $\sqrt\frac{k0 * q1 *q2}{F}$ = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10}$ = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em> $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}$ <em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50}$ = 0,520m

3 0

2 years ago

What creates the changes in step 1,2,3 of the nitrogen cycle

Naya [18.7K]

1-fixation ( Bacteria Converts nitrogen to ammonium so plants can use it )
2-nitrification ( bacteria changes ammonium to nitrates and plants )
3 - Assimilation (plants absorb nitrates it is then used for Chlorophyll..)

6 0

3 years ago

In the context of depth perception, which of the following is a monocular cue?

olga55 [171]

Hello there.

<span>In the context of depth perception, which of the following is a monocular cue?

</span><span>(C) Convergence
</span>

8 0

3 years ago

A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³

Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

= 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

9.5 N = 10 N - weight of water displaced

Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

= 0.051 kg

Therefore the volume of the water displaced,

V = m / ρ

= 0.051 / 1000

= 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

= 5.1 x 10⁻⁵ x 800

= 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

= 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

W₀ = W - w

= 10 N - 0.4 N

= 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0

2 years ago