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3 years ago

8

Four 5-gram blocks of metal are sitting out in the sun and absorb the same amount of heat energy. Use the following specific hea

t capacity data to determine which block will increase its temperature the most.

2 answers:

julsineya [31]3 years ago

5 0

Answer:

There is NO DATA??

Explanation:

Rashid [163]3 years ago

5 0

Answer:copper

Explanation:

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A 2kg mass is connected to a 5kg mass by a non stretchable string.

zaharov [31]

It would be seven kg

7 0

3 years ago

An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?

Luden [163]

Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

3 0

3 years ago

A 3,65 kg mass attached to a

motikmotik

This question involves the concepts of tension, weight, and centripetal force.

The maximum speed, the mass can have before the string breaks is "10.26 m/s".

First, we will find the maximum tension force:

Tension = Weight

T = W = mg = (32.4 kg)(9.81 m/s²)

T = 317.84 N

Now, this tension force must be equal to the centripetal force:

$T = \frac{mv^2}{r}\\v=\sqrt{\frac{Tr}{m}}$

where,

v = maximum speed = ?

r = radius = 1.21 m

m = mass = 3.65 kg

Therefore,

$v=\sqrt{\frac{(317.84\ N)(1.21\ m)}{3.65\ kg}}\\$

<u>v = 10.26 m/s</u>

Learn more about centripetal force here:

brainly.com/question/11324711?referrer=searchResults

The attached picture shows the centripetal force.

8 0

2 years ago

5. A belt runs on a wheel of 30.0-cm radius. During the time that the wheel coasts uniformly to rest from an initial speed of 2.

emmasim [6.3K]

Answer:

Angular acceleration = 0.95rad/s

Number or revolution is 13

Explanation:

See attached file

5 0

3 years ago

A potential difference V = 100 V is applied across a capacitor arrangement with capacitances C1 = 10.0 mF, C2 = 5.00 mF, and C3

Gala2k [10]

Given Information:

Potential difference = V = 100 V

Capacitance C₁ = 10 mF

Capacitance C₂ = 5 mF

Capacitance C₃ = 4 mF

Required Information:

a. Charge q₃

b. Potential difference V₃

c. Stored energy U₃

d. Charge q ₁

e. Potential difference V₁

f. Stored energy U₁

g. Charge q ₂

h. Potential difference V₂

i. Stored energy U₂

Answer:

a. Charge q₃ = 0.4 C

b. Potential difference V₃ = 100 V

c. Stored energy U₃ = 20 J

d. Charge q ₁ = 0.33 C

e. Potential difference V₁ = 33 V

f. Stored energy U₁ = 5.445 J

g. Charge q ₂ = 0.33 C

h. Potential difference V₂ = 66 V

i. Stored energy U₂ = 10.89 J

Explanation:

Please refer to the circuit attached in the diagram

a. Charge q₃

As we know charge in a capacitor is given by

q₃ = C₃V₃

q₃ = 4x10⁻³*100

q₃ = 0.4 C

b. Potential difference V₃

The potential difference V₃ is same as V

V₃ = 100 V

c. Stored energy U₃

Energy stored in a capacitor is given by

U₃ = ½C₃V₃²

U₃ = ½*4x10⁻³*100²

U₃ = 20 J

d. Charge q ₁

Since capacitor C₁ and C₂ are in series their equivalent capacitance is

Ceq = C₁*C₂/C₁ + C₂

Ceq = 10x10⁻³*5x10⁻³/10x10⁻³ + 5x10⁻³

Ceq = 3.33x10⁻³ F

q ₁ = Ceq*V

q ₁ = 3.33x10⁻³*100

q ₁ = 0.33 C

e. Potential difference V₁

V₁ = q ₁/C₁

V₁ = 0.33/10x10⁻³

V₁ = 33 V

f. Stored energy U₁

U₁ = ½C₁V₁²

U₁ = ½*10x10⁻³*(33)²

U₁ = 5.445 J

g. Charge q ₂

q₂ = Ceq*V

q₂ = 3.33x10⁻³*100

q₂ = 0.33 C

h. Potential difference V₂

V₂ = q ₂/C₂

V₂ = 0.33/5x10⁻³

V₂ = 66 V

i. Stored energy U₂

U₂ = ½C₂V₂²

U₂ = ½*5x10⁻³*(66)²

U₂ = 10.89 J

8 0

4 years ago