Ask question

Ask question

All categories

3 years ago

13

Help me! Btw he’s playing golf.

1 answer:

fiasKO [112]3 years ago

5 0

Answer:

Muscular energy

Explanation:

Hope it helps!!!!

You might be interested in

. A student claims that if lighting strikes a metal flagpole, the force exerted by the Earth’s magnetic field on the current in

Ratling [72]

Answer:

Explanation:

Given that, current generated from lightning range from

10⁴ A < I < 10^5 A

We know that,

The magnetic force is given as

F = iLB

The magnetic field on the earth surface is

B = 10^-5 T

So, let assume the worst case of a 15m flag pole

L = 15m

Then,

F = iLB

F = 10^5 × 10 × 10^-5

F = 15 N

Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.

Therefore, it is possible that the student is right depending on the material of the flag pole.

7 0

3 years ago

A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t

Zina [86]

Answer:

The value of spring constant is 266.01 $\frac{N}{m}$

Explanation:

Given:

Mass of pellet $m = 2.01 \times 10^{-2}$ kg

Height difference of pellet rise $h_{f} - h_{o} = 6.03$ m

Spring compression $x = 9.45 \times 10^{-2}$ m

From energy conservation law,

Spring potential energy is stored into potential energy,

$mg(h_{f} -h_{o}) = \frac{1}{2} kx^{2}$

Where $k =$ spring constant, $g = 9.8 \frac{m}{s^{2} }$

$k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }$

$k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }$

$k = 266.01$ $\frac{N}{m}$

Therefore, the value of spring constant is 266.01 $\frac{N}{m}$

6 0

3 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0

3 years ago

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has

vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

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

$a=\frac{13.41-31.29}{5}=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576\times t$

$t=\frac{31.29}{3.576}=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2\times (-3.576)\cdot s$

s=136.89 m

3 0

3 years ago

A 10,000 N net force is accelerating a car at a rate of 5.5m/s^2. What is the cars mass

ycow [4]

Answer:

If a net horizontal force of 175 N is applied to a bike whose mass is 43 kg what acceleration is produced? What average net force is ... A 10,000 N net force is accelerating a car at a rate of 5.5 m/s2. What is the car's mass? A boy pedals his ...

Explanation:

6 0

3 years ago