Ask question

Ask question

All categories

3 years ago

11

One of Lex Luthor's henchman attacks Superman, shooting a rapid-fire stream of 3.3 g bullets at him at a rate of 112/min. The sp

eed of each bullet is 527 m/s. Because of his super-power, the bullets just stop and fall straight to the ground after striking his chest. What is the magnitude of the average force on Superman's chest?

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

3.2451N

Explanation:

Mass of the bullet (m) $= 3.3g = 3.3*10^{-3}Kg$

Speed of the bullet (V) $= 527m/s$

Rate of bullet (r) $= 112/min = 1.866\sec$

We can calculate with this information the average acceleration of bullets

$a=V*r = 527\frac{m}{s}\frac{1.866}{s} = 983.38m/s^2$

The force is given by,

$F=ma\\F=(3.3*10^{-3})*983.38m/s^2 = 3.2451N$

That is just because he is Superman.

You might be interested in

Describe the three particles that make up an atom (name, charge, and location in the atom).

kow [346]

Answer:Protons, neutrons, and electrons are the three main subatomic particles found in an atom. Protons have a positive (+) charge. An easy way to remember this is to remember that both proton and positive start with the letter "P." Neutrons have no electrical charge.

5 0

3 years ago

Read 2 more answers

Carbon and hydrogen are examples of pure substances or

Ede4ka [16]

The are elements on the periodic table

7 0

4 years ago

Read 2 more answers

A vector consists both?

Diano4ka-milaya [45]

Answer:

magnitude and a direction

Explanation:

3 0

3 years ago

ANSWER QUICKLY PLEASE <3

Sladkaya [172]

The forces on one of the wires will be 0.000264 N.It is the field sensed around a moving electric charge that gives rise to the magnetic force.

<h3>What is a magnetic field?</h3>

It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.

The given data in the problem is;

I is the current = 6.50 A

B is the magnetic field=f 2.71 X 10⁻⁵ T

The force on the wire is;

$\rm F= BIL \\\\\ F=2.71 \times 10^{-5} \ T \times 6.50 \ A \times 1.50 \ m \\\\\ F=0.000264225 \ N$

Hence, the force on one of the wires will be 0.000264 N.

To learn more about the magnetic field refer to the link;

brainly.com/question/19542022

#SPJ1

6 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago