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Ipatiy [6.2K]
3 years ago
14

If the heat necessary to warm 565.0 g of water from a temperature of T1 = 22.0 °C to T2 = 80.0 °C were somehow converted to tran

slational kinetic energy of this amount of water, what would be the speed of this water?
Physics
1 answer:
densk [106]3 years ago
3 0

Answer:

696.83 m/s

Explanation:

m = mass of water = 565 g = 0.565 kg

c = specific heat of water = 4186 J/(kg⁰C)

ΔT = Change in temperature = T₂ - T₁ = 80 - 22 = 58 ⁰C

v = speed gained by water

Using conservation of energy

Kinetic energy gained by water = heat required to warm water

(0.5) m v² = m c ΔT

(0.5) v² = c ΔT

(0.5) v² = (4186) (58)

v = 696.83 m/s

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Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t
LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

F_{c} = \frac{mv^{2}}{R}  

f_{s} = \mu_{s}mg

where

\mu_{s} = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

F_{net} = f_{s} - F_{c}  

0 = f_{s} - F_{c}  

\mu_{s}mg = \frac{mv^{2}}{R}  

\mu_{s} = \frac{v^{2}}{gR}  

\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544  

3 0
4 years ago
A ball was positioned in the middle of a smooth ramp and allowed to roll
netineya [11]

Answer:

it is quicker

Explanation:

6 0
2 years ago
Proposed Exercise - Circular Movement
notka56 [123]

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

5 0
3 years ago
A ship sets out to sail to a point 154 km due north. An unexpected storm blows the ship to a point 72 km due east of its startin
Norma-Jean [14]

Answer:

<h2>170km</h2>

Explanation:

If a ship sets out to sail to a point 154 km due north and an unexpected storm blows the ship to a point 72 km due east of its starting point, then the ships distance from the original destination can be gotten by finding the displacement of the ship and this can be gotten by using pythagoras theorem.

Let D be the unknown displacement

According to the theorem;

D² = 154² + 72²

D² = 23716 + 5184

D² = 28900

D = √28900

D = 170km

<em>This means that the ship must now sail a distance of 170km for it to reach its original destination.</em>

8 0
3 years ago
Federal board paper of class 10 of physics​
zzz [600]

Answer:

If you are looking for past papers you can search that up and you will find plenty of resources that will help you out.

8 0
3 years ago
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