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DanielleElmas [232]

3 years ago

13

A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after

it has traveled 100 meters?

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

125,000

Explanation:

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A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

Learn More:

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With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward brainly.com/question/9719731
Which of the following is an example of a nonpoint source of freshwater pollution brainly.com/question/1482712

Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

Read 2 more answers

True or false cells of humans and plants are same.

timama [110]

False, we lack cell walls whereas they have cell walls.

4 0

2 years ago

Electrical current is defined as

Dmitry_Shevchenko [17]

Explanation:

The flow of electrical charge per unit time

8 0

2 years ago

The paths of the alpha particles depend on the forces on them in the metal.

grandymaker [24]

Answer:

During the experiment, alpha particles bombarded a thin piece of gold foil. The alpha particles were expected to pass easily through the gold foil. ... The discovery of the nucleus was a result of Rutherford's observation that a small percentage of the positively charged particles bombarding the metal's surface...

Explanation:

jayfeather friend me

6 0

2 years ago

An electron travels 0.01 m across an electric potential difference of 6 x 10^6 V. What is the force exerted on the electron? Typ

JulijaS [17]

Answer:

$9.6\cdot 10^{-11}N$

Explanation:

The force experienced by a charged particle immersed in an electric field is

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

For a uniform field, we can write

$E=\frac{\Delta V}{d}$

where

$\Delta V$ is the potential difference

d is the distance travelled

So we can rewrite the force as

$F=\frac{q\Delta V}{d}$

In this problem:

$q=1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=6\cdot 10^6 V$ is the potential difference

d = 0.01 m is the distance covered

Substituting, we find the force:

$F=\frac{(1.6\cdot 10^{-19})(6\cdot 10^6)}{0.01}=9.6\cdot 10^{-11}N$

3 0

2 years ago