How can I solve this?

2.0 kg mass is lifted4.0m above the ground find the gravitational energy

melisa1 [442]

E = mgh
= (2)(9.8)(4)
= 78.4 Joules

3 0

3 years ago

Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle. (Recall, the density of wate

xeze [42]

Answer:

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

Explanation:

Given;

Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle

Initial speed v1 = 0.65 m/s

diameter d1 = 3.0 cm

diameter (nozzle) d2 = 0.3 cm

The volumetric flow rates in both the hose and the nozzle are the same.

V1 = V2 ........1

Volumetric flow rate V = cross sectional area × speed of flow

V = Av

Area = (πd^2)/4

V = v(πd^2)/4 ....2

Substituting equation 2 to 1;

v1(πd1^2)/4 = v2(πd2^2)/4

v1d1^2 = v2d2^2

v2 = (v1d1^2)/d2^2

Substituting the given values;

v2 = (0.65 × 3^2)/0.3^2

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

6 0

4 years ago

if you transfer equal amounts of geat to 1kg of water and 1kg of cooper, which will change tempature more

Angelina_Jolie [31]

If you transfer equal amounts of geat to 1kg of water and 1kg of cooper, the temperature of the cooper will change more . . . it'll get more degrees warmer than the water will.

That's because the specific geat of water is greater than the specific geat of cooper.

(This just means it takes more geat to warm some mass of water by some amount than it takes to warm the same mass of cooper by the same amount.)

4 0

3 years ago

3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is

miskamm [114]

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}$

Applying values we get

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)$

4 0

3 years ago

Read 2 more answers

Physics !!!!!!!!!!!!!!!!!

madreJ [45]

Answer:

1) v = -0.89 m/s

2) v = 36.5 m/s

Explanation:

1) Given

The mass of the boy, M = 45 Kg

The mass of the skateboard, m = 2 Kg

The initial velocity of the boy and skateboard is zero

According to the law of conservation of linear momentum

MV + mv = 0

V = - mv/M

Substituting the given values in the above equation

V = -2 x 20/ 45

= -0.89 m/s

Hence, the velocity of the boy, v = -0.89 m/s

2) Given

The mass of the boy, M = 47 Kg

The mass of the skateboard, m = 8 Kg

The initial combined velocity of the boy and skateboard, u = 4.2 m/s

The boy jumped off from the skateboard with velocity, V = -1.3 m/s

According to the law of conservation of momentum,

MV + mv = (M + m)u

v = [(M + m)u - MV] / m

Substituting the given values,

v = [(47 + 8)4.2 - 47(-1.3)] / 8

= 36.5 m/s

Hence, the velocity of the skateboard, v = 36.5 m/s

3 0

4 years ago