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3 years ago

Answer these questions with step by step

Physics

2 answers:

Harlamova29_29 [7]3 years ago

6 0

Answer:

i dont rlY KNow sorryyyyyyyyy

Explanation:

DedPeter [7]3 years ago

5 0

I’m pretty sure you should just guess cause

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The acceleration due to gravity on the surface of Venus is 8.83 m/s2. An object with a mass of 5.23 kg has what weight on Venus?

Marat540 [252]

Weight = m times g = 5.23 times 8.83 = 46.18 N

6 0

3 years ago

A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. a. centimeters per hour (cm/h) b. c

guajiro [1.7K]

It would be 300 cm/h
and 5 cm/m

4 0

3 years ago

Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd

Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0

3 years ago

When does a rubber band, which has been shot at a wall, have the most potential energy?

Yuri [45]

D When it is stretched ready to shoot at the wall

4 0

2 years ago

Read 2 more answers

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

3 years ago