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2 years ago

Please please help I'm stuck!!!

Physics

2 answers:

alex41 [277]2 years ago

8 0

I GOTCHU THE ANSWER IS 2.44s

Korvikt [17]2 years ago

6 0

Answer: iDc im in 3grade

Explanation:sorry

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A relatively small impact crater 20 kilometers in diameter could be made by a comet 2 kilometers in diameter traveling at 25 km/

nadya68 [22]

Answer:

$KE=1.3\times 10^{21}\ J$

$KE=3.1\times 10^{5}\ E$

Explanation:

(a)

Given:

mass of comet, $m=4.2\times 10^{12}\ kg$

velocity of the comet, $v=2.5\times 10^4\ m.s^{-1}$

<u>Now, the kinetic energy of the comet can be given by:</u>

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 4.2\times 10^{12}\times (2.5\times 10^4)^2$

$KE=1.3\times 10^{21}\ J$

(b)

Given:

energy released by 1 megaton of TNT, $E=4.2\times 10^{15}\ J$

<u>Now the kinetic energy of the comet in terms of energy of 1 megaton TNT:</u>

$KE=\frac{1.3\times 10^{21}}{4.2\times 10^{15}} E$

i.e.

$KE=3.1\times 10^{5}\ E$

4 0

3 years ago

A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determin

Paul [167]

Answer:

$F_x=208.25\ N$

Explanation:

Given that,

Mass of a crate is 22 kg

It moved up along the 15 degrees incline without tipping.

We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.

It means that the horizontal component of force is given by :

$F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N$

So, the horizontal component of force is 208.25 N.

6 0

3 years ago

A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c

satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

Given the data in the question;

First course : $a_{x}$ = 0.75 m/s², $d_{x}$ = 20 m, $u_{x}$ = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

$v_{x}$ ² = (10)² + (2 × 0.75 × 20)

$v_{x}$ ² = 100 + 30

$v_{x}$ ² = 130

$v_{x}$ = √130

$v_{x}$ = 11.4 m/s

for the Second Course:

$u_{y}$ = 11.4 m/s, $a_{y}$ = -1.15 m/s², $v_{y}$ = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) × $d_{y}$ )

0 = 129.96 - 2.3 $d_{y}$

2.3 $d_{y}$ = 129.96

$d_{y}$ = 129.96 / 2.3

$d_{y}$ = 56.5 m

so;

|d| = √( $d_{x}$ ² + $d_{y}$ ² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0

2 years ago

How does the energy possessed by a ball bearing change as it travels along an incline ramp?

ruslelena [56]

Kinetic energy increases, potencial energy decreases,
kinetic energy + potential energy = energy, energy can not be destroyed, just transformed

7 0

3 years ago

A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$

$N_2 = 1400$

$\Delta V_{1rms} = 62V$

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

$\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}$

So the voltage for transformer 2 would be given by,

$\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}$

PART B) To express the number value we proceed to replace with the previously given values, that is to say

$\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}$

$\Delta V_{2rms} = \frac{1400}{160} 62V$

$\Delta V_{2rms} = 1446.66V$

7 0

3 years ago