Which type of boundary is modeled?

A sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 1

hammer [34]

The specific heat of the substance will be 0.129 J/g°C.

<h3>What is specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The given data in the problem is;

Q is the amount of energy necessary to raise the temperature = 3,000.0 j

M is the mass= 0.465 kg.

Δt is the time it takes to raise the temperature.=50°c

s stands for specific heat capacity=?

Mathematically specific heat capacity is given by;

$\rm Q= MC \triangle t \\\\ C = \frac{Q}{M\triangle t} \\\\ C = \frac{3000}{0.465 \triangle 50} \\\\ C =129.0 J/Kg^0C \\\\ C= 0.129 J/g^0C$

Hence the specific heat of the substance will be 0.129 J/g°C.

To learn more about the specific heat capacity refer to the link brainly.com/question/2530523

5 0

2 years ago

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

Read 2 more answers

A force of 30.0 N is applied to a 3.00 kg object for 3.00 seconds. Calculate the velocity experienced by the object.

olganol [36]

Answer:

Explanation:

F = ma and

$a=\frac{v}{t}$