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3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (47pts)

Physics

2 answers:

Reptile [31]3 years ago

5 0

Answer:

can I have points, i don't have any money

Airida [17]3 years ago

4 0

gave you earlier :) Check

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What is Yusef most likely finding?

Ratling [72]

Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

4 0

3 years ago

the ball is accelerating at a rate of 10 m/s squared. the balls mass if 5 kg. what is the net force on the ball? help me men ple

nevsk [136]

Answer:

50N

Explanation:

$F=ma\\F=(5kg)*(10m/s^{2})\\F=50N$

6 0

2 years ago

you kick a soccer ball straight up into the air with a speed of 21.2 m/s how long does it take the soccer ball to reach its high

skelet666 [1.2K]

i just had this question and i totally guessed. it was 2.2.

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3 years ago

If a planet has 3 times the radius of the Earth, but has the same density as the Earth, what is the gravitational acceleration a

xz_007 [3.2K]

Answer:

The Gravitational Acceleration of the Big planet is G = 3g

where G is the gravitational acceleration of the Big Planet and g is the gravitational acceleration of the Earth

So, it becomes G = 3 x 9.8 $ms^{-2}$

G = 29.4 $ms^{-2}$

Explanation:

Let's start by calculating the <u>Density of the Earth</u> of mass m anad Radius r

We know that mass is density times the volume

ρ = $\frac{m}{V}$

m = ρV

we know the volume is V = $\frac{4}{3}$ $πr^{3}$ (please ignore the symbol of Pi)

m = ρ $\frac{4}{3}$ $πr^{3}$

Calculating the <u>Density of Big Planet</u>

→For Big Planet we know that radius is 3-times the radius of Earth, that is 3r

M = ρ $\frac{108}{3}$ $π(r)^{3}$

So in conclusion, the mass of Earth and the mass of Big planet are related as M = 27m

Now let's come to the Gravitational Force, we know that gravitational force is directly proportional to the mass of the body and inversely proportional to the radius.

→Suppose for Earth: Mass = m, Radius = r

For Big Planet: Mass = M, Radius = R

$\frac{m}{r^{2} }$ : $\frac{M}{R^{2} }$ = g : G

$\frac{Gm}{r^{2} }$ = $\frac{gM}{R^{2} }$

G = $\frac{gMr^{2}}{mR^{2}}$

→ Putting the value of R = 3r and M = 27m

G = $\frac{g27mr^{2}}{m(3r)^{2}}$

By cutting the terms, we get

G = 3g

value of g= 9.8 $ms^{-2}$

G = 3 x $9.8ms^{-2}$

G = 29.4 $ms^{-2}$

5 0

4 years ago

A radioisotope is place near a radiation detector, which registers 80 counts per second. Eight hours later, the detector registe

vesna_86 [32]

The isotopes half life is 2 hours

Since given the following :
time = 8 hours × 3600s/hr = 28800 seconds
A₀ = 80 Bq as the first registration of detector counts per second
A = 5 Bq as the registration of detector counts per second eight hours later

Using the formula:
∴ A = A₀ e^-λt
∴ A/A₀ = e^-λt
∴ ln (A/A₀) = -λt
∴ λ = -ln (A/A₀) /t = ln2 / t_{1/2}
∴ t_{1/2} = -ln2 / ln(A/A₀) (t)
∴ t_{1/2} = -ln2 / ln(5/80) (8 hours) = 2 hours

5 0

3 years ago