6. a steel ball whose mass is 2 kg is rolling at a rate of 2.8 m/s. what is its momentum?
7. a marble is rolling at a velocity of 1.5 m/s with a momentum of 0.10 kg(m/s). what is its mass?
8. find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground?
9. a marble is on a table 2.4 m above the ground. what is the mass of the marble if it has a gravitational potential energy of 568 j?
10. what is the kinetic energy of a jogger with a mass of 65.0 kg traveling at a speed of 2.5 m/s?
11. what is the mass of a baseball that has a kinetic energy of 105 j and is traveling at 10 m/s?
12. how much heat must be absorbed by 375 grams of water to raise its temperature by 25°c?

Question

Accepted Answer

### Problem 6
# Explanation:
## Step1: Recall momentum formula
The formula for momentum $ p $ is $ p = mv $, where $ m $ is mass and $ v $ is velocity.
## Step2: Substitute values
Given $ m = 2 \, \text{kg} $ and $ v = 2.8 \, \text{m/s} $, substitute into the formula: $ p = 2 \times 2.8 $.
## Step3: Calculate
$ 2 \times 2.8 = 5.6 \, \text{kg·m/s} $.
# Answer:
The momentum of the steel ball is $ 5.6 \, \text{kg·m/s} $.

### Problem 7
# Explanation:
## Step1: Recall momentum formula
The formula for momentum is $ p = mv $, so we can solve for mass $ m $ as $ m=\frac{p}{v} $.
## Step2: Substitute values
Given $ p = 0.10 \, \text{kg·m/s} $ and $ v = 1.5 \, \text{m/s} $, substitute into the formula: $ m=\frac{0.10}{1.5} $.
## Step3: Calculate
$ \frac{0.10}{1.5}=\frac{1}{15}\approx0.067 \, \text{kg} $ (or $ 67 \, \text{g} $).
# Answer:
The mass of the marble is approximately $ 0.067 \, \text{kg} $ (or $ 67 \, \text{g} $).

### Problem 8 (Assuming $ g = 9.8 \, \text{m/s}^2 $ and the object is a "light" (maybe a typo, likely "light - object") with $ m = 13.0 \, \text{kg} $, $ h = 4.8 \, \text{m} $)
# Explanation:
## Step1: Recall gravitational potential energy formula
The formula for gravitational potential energy $ PE_g $ is $ PE_g = mgh $, where $ g = 9.8 \, \text{m/s}^2 $, $ m $ is mass, and $ h $ is height.
## Step2: Substitute values
Given $ m = 13.0 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $, and $ h = 4.8 \, \text{m} $, substitute into the formula: $ PE_g=13.0\times9.8\times4.8 $.
## Step3: Calculate
$ 13.0\times9.8 = 127.4 $, then $ 127.4\times4.8 = 611.52 \, \text{J} $.
# Answer:
The gravitational potential energy is $ 611.52 \, \text{J} $.

### Problem 9
# Explanation:
## Step1: Recall gravitational potential energy formula
The formula for gravitational potential energy is $ PE_g = mgh $, so we can solve for mass $ m $ as $ m=\frac{PE_g}{gh} $.
## Step2: Substitute values
Given $ PE_g = 568 \, \text{J} $, $ g = 9.8 \, \text{m/s}^2 $, and $ h = 2.4 \, \text{m} $, substitute into the formula: $ m=\frac{568}{9.8\times2.4} $.
## Step3: Calculate
First, $ 9.8\times2.4 = 23.52 $, then $ \frac{568}{23.52}\approx24.15 \, \text{kg} $.
# Answer:
The mass of the marble is approximately $ 24.15 \, \text{kg} $.

### Problem 10
# Explanation:
## Step1: Recall kinetic energy formula
The formula for kinetic energy $ KE $ is $ KE=\frac{1}{2}mv^2 $, where $ m $ is mass and $ v $ is speed.
## Step2: Substitute values
Given $ m = 65.0 \, \text{kg} $ and $ v = 2.5 \, \text{m/s} $, substitute into the formula: $ KE=\frac{1}{2}\times65.0\times(2.5)^2 $.
## Step3: Calculate
First, $ (2.5)^2 = 6.25 $, then $ \frac{1}{2}\times65.0\times6.25 = 32.5\times6.25 = 203.125 \, \text{J} $.
# Answer:
The kinetic energy of the jogger is $ 203.125 \, \text{J} $.

### Problem 11
# Explanation:
## Step1: Recall kinetic energy formula
The formula for kinetic energy is $ KE=\frac{1}{2}mv^2 $, so we can solve for mass $ m $ as $ m = \frac{2KE}{v^2} $.
## Step2: Substitute values
Given $ KE = 105 \, \text{J} $ and $ v = 10 \, \text{m/s} $, substitute into the formula: $ m=\frac{2\times105}{(10)^2} $.
## Step3: Calculate
First, $ 2\times105 = 210 $, then $ (10)^2 = 100 $, so $ \frac{210}{100}=2.1 \, \text{kg} $.
# Answer:
The mass of the baseball is $ 2.1 \, \text{kg} $.

### Problem 12
# Explanation:
## Step1: Recall heat formula for temperature change
The formula for heat $ Q $ absorbed is $ Q = mc\Delta T $, where $ m $ is mass, $ c $ is specific heat capacity (for water, $ c = 4.186 \, \text{J/g·°C} $ or $ 4186 \, \text{J/kg·°C} $), and $ \Delta T $ is temperature change. Given $ m = 375 \, \text{g} $, $ \Delta T = 25 \, \text{°C} $, and $ c = 4.186 \, \text{J/g·°C} $.
## Step2: Substitute values
Substitute into the formula: $ Q = 375\times4.186\times25 $.
## Step3: Calculate
First, $ 375\times4.186 = 1570.5 $, then $ 1570.5\times25 = 39262.5 \, \text{J} $ (or $ 39.2625 \, \text{kJ} $).
# Answer:
The heat absorbed by the water is $ 39262.5 \, \text{J} $ (or $ 39.26 \, \text{kJ} $).

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QUESTION IMAGE

Question

Explanation:

Problem 6

Step1: Recall momentum formula

Step2: Substitute values

Step3: Calculate

Step1: Recall momentum formula

Step2: Substitute values

Step3: Calculate

Step1: Recall gravitational potential energy formula

Step2: Substitute values

Step3: Calculate

Answer:

Problem 7