if a 10kg object is lifted a distance of 10m from the rest what was the final velocity of the object?

A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the g

soldi70 [24.7K]

Answer:

I think answer is zero

bcz momentum=mass×velocity

body was initially at rest it means its velocity is zero

30×0=0

6 0

3 years ago

How many units measure one wavelength? a 16 b 8 c 2 d 4

pochemuha

B it makes more sense

8 0

3 years ago

A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga

tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₂² v₂

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{2.25^2}{3.75^2} v_2$

$v_1= 0.36 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)$

$32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)$

$v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}$

$v_2=\sqrt{16.084}$

v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)² x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0

3 years ago

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

Paladinen [302]

Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

8 0

3 years ago

A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t

marishachu [46]

Answer:

t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

864 = 240 t + 16 t²

Let's solve the second degree equation

16 t² + 240 t - 864 = 0

Let's divide by 16

t² + 15 t - 54 = 0

The solution of this equation is

t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

t = [-15 ±√(225 +216)] / 2

t = [-15 + - 21] / 2

We have two solutions.

t₁ = [-15 +21] / 2

t₁ = 3 s

t₂ = -18 s

Since time cannot have negative values, the correct t₁ = 3s

4 0

3 years ago