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skelet666 [1.2K]
3 years ago
6

If the specific heat of water is 1.0 calorie/gram°C, how many calories are required to raise 500 grams of water 10.0°C?

Physics
1 answer:
kykrilka [37]3 years ago
6 0
The amount of heat that is required to raise the temperature of a system is calculated by the expression:

Heat required = m Cp ΔT
Heat required = 500 g (1.0 calorie/gram °C) (10.0 <span>°C)
Heat required = 5000 calorie
 
The correct answer from the choices listed above is the second option.</span>
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A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i
Stolb23 [73]

The speed  V_{i} of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.

We are given that-

the mass of the elevator (m) = 1000 kg ;

the distance the elevator decelerated to be y = 8m ;

the tension is T = 11000 N;

let us determine the acceleration 'a' by using Newton's second law of motion.

∑Fy = ma

W - T = ma

(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a

9800 - 11000 = 1000

a = - 1.2 m/s²

Using the equation of kinematics to determine the initial velocity.

V_{f} ² = V_{i}² + 2ay

V_{i} = √ ( 2 x 1.2m/s² x 8 m )

V_{i} = √19.2 m²/s²

V_{i} = 4.38 m/s   ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

Read more about the Equation of kinematics:

brainly.com/question/12351668

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8 0
1 year ago
A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an
allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

K_i + U_i = K_f + U_f

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K =  (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×v_f² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75×v_f²

v_f² = 77.93/(7/10 × 4.75)

v_f ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, v_f ≈ 4.84 m/s.

3 0
2 years ago
A runner first runs a displacement A of 3.20 km due south, and then a second displacement B that points due east. (a) The magnit
Doss [256]

Answer: 4,438.96m

Explanation:

(kindly find attachment below)

From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.

By using phythagoras theorem

H² = O² + A²

(5.38)² = (3.20)² + B²

28.944 = 10.24 + B²

B² = 28.944 - 10.24

B² = 18.7044

B = √18.7044

B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m

3 0
3 years ago
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Kate must do 39 days of training each year as a naval reservist. How will she serve that time?
elixir [45]

Answer:

c

Explanation:

i did the test

3 0
3 years ago
Which advantage of stored digital data could be a disadvantage in some
Lesechka [4]

Answer:B- All types of information can be pressed on to receivers of the data.

Explanation: hope this helps

4 0
3 years ago
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