A hiker walks 7.45 m, N and 2.50 m. E. What is the magnitude of his resultant

This graph shows velocity vs. time. What does the slope of the line represent? A. speed B. force C. acceleration D. distance

PilotLPTM [1.2K]

I believe the answer is C. Acceleration. (:

8 0

4 years ago

What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011

nlexa [21]

Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

<h3>What is electric force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.

$\rm F= K\frac{q_2q_3}{r^2}$

$\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}$

$\rm F= - 1. 1 \times 10^{11 }N$

Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

To learn more about the electric force refer to the link;

brainly.com/question/1076352

4 0

3 years ago

A pendulum makes 60 vibrations in 15 secs what’s its frequency

o-na [289]

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

so,

cycle or vibrations=60

tame taken = 15

now,

frequency = no. of cycle / time taken

= 60/15

=4 hertz

hence, the its frequency = 4hertz

8 0

4 years ago

What are ways to improve the design of this experiment? Check all that apply

Serhud [2]

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.

E. Test the strength of the electromagnet by varying the number of wire coils.

3 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago