A stone of mass 10 kg . is suspended by a rope that can not afford tension more than 30 . , then the least time needed to ciise the stone 20 m . vertically upward from rest equals second.
(a)
(b)
© 2
(d)

A bullet is fired horizontally with speed . at a vertical fixed barrier of thickness 32 cm . it penetrates the barrier and lose of its speed If the barrier resistance ., then the mass of the bullet = gm.
(a) 5
(b) 10
© 12
(d) 15

A bullet of mass 7 gm . is fired horizontally from a pistol with speed . at a wooden vertical barrier. It imbedded in it 12.25 cm . befor it stopped, then the resistance of the wood to the bullet equals
(a) 17.15 N
(b) 175 N .
© .
(d) .

Ask by Peterson Vargas. in Egypt

Mar 27,2025

Question

A stone of mass 10 kg . is suspended by a rope that can not afford tension more than 30 . , then the least time needed to ciise the stone 20 m . vertically upward from rest equals second.
(a)
(b)
© 2
(d)

A bullet is fired horizontally with speed . at a vertical fixed barrier of thickness 32 cm . it penetrates the barrier and lose of its speed If the barrier resistance ., then the mass of the bullet = gm.
(a) 5
(b) 10
© 12
(d) 15

A bullet of mass 7 gm . is fired horizontally from a pistol with speed . at a wooden vertical barrier. It imbedded in it 12.25 cm . befor it stopped, then the resistance of the wood to the bullet equals
(a) 17.15 N
(b) 175 N .
© .
(d) .

Ask by Peterson Vargas. in Egypt

Mar 27,2025

UpStudy AI · Accepted Answer

Let's solve each problem step by step.

### Problem 1: Stone Suspended by a Rope

**Given:**
- Mass of the stone, $$ m = 10 \, \text{kg} $$
- Maximum tension the rope can afford, $$ T_{\text{max}} = 30 \, \text{kg} \cdot \text{m/s}^2 $$ (which is equivalent to $$ 30 \, \text{N} $$)
- Distance to be raised, $$ d = 20 \, \text{m} $$

**Step 1: Calculate the net force acting on the stone.**
The maximum tension in the rope must overcome both the weight of the stone and provide the necessary force to accelerate it upwards.

The weight of the stone is given by:
$$
W = m \cdot g = 10 \cdot 9.8 = 98 \, \text{N}
$$

The net force $$ F_{\text{net}} $$ can be expressed as:
$$
F_{\text{net}} = T - W
$$
Where $$ T $$ is the tension in the rope.

**Step 2: Set up the equation for maximum tension.**
Since the maximum tension is $$ 30 \, \text{N} $$:
$$
30 = F_{\text{net}} + 98
$$
This implies:
$$
F_{\text{net}} = 30 - 98 = -68 \, \text{N}
$$
This indicates that the tension is not sufficient to lift the stone.

**Step 3: Calculate the acceleration.**
Using Newton's second law:
$$
F_{\text{net}} = m \cdot a
$$
We can rearrange this to find $$ a $$:
$$
a = \frac{F_{\text{net}}}{m} = \frac{-68}{10} = -6.8 \, \text{m/s}^2
$$
This negative acceleration indicates that the stone cannot be lifted with the given tension.

Since the stone cannot be lifted, we cannot calculate the time needed to raise it.

### Problem 2: Bullet Fired at a Barrier

**Given:**
- Initial speed of the bullet, $$ v_0 = 200 \, \text{m/s} $$
- Thickness of the barrier, $$ d = 0.32 \, \text{m} $$
- Speed loss, $$ \frac{4}{5} $$ of its speed
- Resistance of the barrier, $$ R = 900 \, \text{N} $$

**Step 1: Calculate the final speed of the bullet after penetrating the barrier.**
The final speed $$ v_f $$ is:
$$
v_f = v_0 - \frac{4}{5} v_0 = \frac{1}{5} v_0 = \frac{1}{5} \cdot 200 = 40 \, \text{m/s}
$$

**Step 2: Calculate the deceleration.**
Using the formula:
$$
v_f^2 = v_0^2 + 2a d
$$
Rearranging gives:
$$
a = \frac{v_f^2 - v_0^2}{2d}
$$
Substituting the values:
$$
a = \frac{40^2 - 200^2}{2 \cdot 0.32}
$$

**Step 3: Calculate the mass of the bullet.**
Using Newton's second law:
$$
F = m \cdot a
$$
Where $$ F = R $$:
$$
900 = m \cdot a
$$
We will calculate $$ a $$ first and then find $$ m $$.

### Problem 3: Bullet Embedded in Wood

**Given:**
- Mass of the bullet, $$ m = 7 \, \text{g} = 0.007 \, \text{kg} $$
- Initial speed, $$ v_0 = 245 \, \text{m/s} $$
- Distance embedded, $$ d = 0.1225 \, \text{m} $$

**Step 1: Calculate the deceleration.**
Using the same formula as before:
$$
v_f^2 = v_0^2 + 2a d
$$
Since the bullet stops, $$ v_f = 0 $$:
$$
0 = 245^2 + 2a \cdot 0.1225
$$
Rearranging gives:
$$
a = -\frac{245^2}{2 \cdot 0.1225}
$$

**Step 2: Calculate the resistance of the wood.**
Using $$ F = m \cdot a $$:
$$
F = 0.007 \cdot a
$$

Now, let's calculate the values for Problems 2 and 3.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{245^{2}}{\left(2\times 0.1225\right)}$$
- step1: Remove the parentheses:
  $$\frac{245^{2}}{2\times 0.1225}$$
- step2: Multiply the numbers:
  $$\frac{245^{2}}{0.245}$$
- step3: Convert the expressions:
  $$\frac{245^{2}}{\frac{49}{200}}$$
- step4: Evaluate the power:
  $$\frac{60025}{\frac{49}{200}}$$
- step5: Multiply by the reciprocal:
  $$60025\times \frac{200}{49}$$
- step6: Reduce the numbers:
  $$1225\times 200$$
- step7: Multiply the numbers:
  $$245000$$
Calculate or simplify the expression $$ (40^2 - 200^2)/(2*0.32) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(40^{2}-200^{2}\right)}{\left(2\times 0.32\right)}$$
- step1: Remove the parentheses:
  $$\frac{40^{2}-200^{2}}{2\times 0.32}$$
- step2: Multiply the numbers:
  $$\frac{40^{2}-200^{2}}{0.64}$$
- step3: Subtract the numbers:
  $$\frac{-38400}{0.64}$$
- step4: Convert the expressions:
  $$\frac{-38400}{\frac{16}{25}}$$
- step5: Multiply by the reciprocal:
  $$-38400\times \frac{25}{16}$$
- step6: Reduce the numbers:
  $$-2400\times 25$$
- step7: Multiply the numbers:
  $$-60000$$
Calculate or simplify the expression $$ 0.007*245000 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$0.007\times 245000$$
- step1: Multiply the numbers:
  $$1715$$
Calculate or simplify the expression $$ 900/(-60000) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{900}{\left(-60000\right)}$$
- step1: Remove the parentheses:
  $$\frac{900}{-60000}$$
- step2: Reduce the fraction:
  $$\frac{3}{-200}$$
- step3: Rewrite the fraction:
  $$-\frac{3}{200}$$
### Problem 1: Stone Suspended by a Rope
The stone cannot be lifted with the given tension of 30 N, as the weight of the stone (98 N) exceeds the maximum tension. Therefore, we cannot calculate the time needed to raise the stone.

### Problem 2: Bullet Fired at a Barrier
1. **Deceleration**: 
   $$
   a = -60000 \, \text{m/s}^2
   $$
2. **Mass of the bullet**:
   $$
   m = \frac{900}{60000} = 0.015 \, \text{kg} = 15 \, \text{g}
   $$
   Thus, the mass of the bullet is **15 g**.

### Problem 3: Bullet Embedded in Wood
1. **Deceleration**:
   $$
   a = 245000 \, \text{m/s}^2
   $$
2. **Resistance of the wood**:
   $$
   F = 0.007 \cdot 245000 = 1715 \, \text{N}
   $$
   Thus, the resistance of the wood to the bullet is **1715 N**.

### Summary of Answers:
1. **Stone Problem**: Cannot be lifted.
2. **Bullet Mass**: 15 g
3. **Resistance of Wood**: 1715 N
Cannot lift the stone with the given tension. The bullet's mass is 15 grams, and the wood's resistance is 1715 Newtons.

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