All categories

2 years ago

What is the temperature increase of 4.0 kg of water when it is heated by

Physics

1 answer:

Nikolay [14]2 years ago

6 0

Answer:

ΔT = 28.667°C is the answer.

Explanation:

Given, Mass of water = 4.0 kg

Water is heated by = 8 × 10² W = 800 W

Time = 10 min = 10 × 60 = 600 s

Now, using the equation of specific heat,

Q = mcΔT

where c is the specific heat capacity of water = 4186 J/kg°C

Q = Pt

Substituting,

Pt = mcΔT

800 × 600 = 4 × 4186 × ΔT

480000 = 16744 × ΔT

ΔT = 28.667

Temperature increased, ΔT = 28.667°C

You might be interested in

Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be

ANTONII [103]

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

5 0

3 years ago

A ball is rolled uphill a distance of 12 meters before it slows, stops, and begins to roll back. The ball rolls downhill 20 mete

Sedaia [141]

The ball rolled a distance of
d = 12m + 20m.
But the change of position is
x = + 12m - 20m

5 0

3 years ago

How can you describe the motion of an object in a race?

Hitman42 [59]

You can describe the motion of an object by its position, speed, direction, and acceleration

4 0

2 years ago

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.

SpyIntel [72]

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by $\tau =F\times d=45.5\times 0.0245=1.11475N-m$

So torque on the rocket will be 1.11475 N -m

3 0

3 years ago

The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie

pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

$\bf \longrightarrow {v}^{2} = {u}^{2} + 2as \\ \\ \rm \longrightarrow {0}^{2} = {15}^{2} + 2 \times a \times 63 \\ \\ \rm \longrightarrow 0 = 225 + 126a \\ \\ \rm \longrightarrow 126a = - 225 \\ \\ \rm \longrightarrow a = - \dfrac{225}{126} \\ \\ \rm \longrightarrow a = - 1.78 \: m {s}^{ - 2}$

$\therefore$ Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0

3 years ago