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3 years ago

What is the similarity between relative dating and radioactive dating?

1 answer:

4 0

They both provide a range of years of an object. I think. They’re just 2 different ways to tell the age of fossils or rocks

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What average net force is required to stop a 4.5 kg bowling ball,initially at rest, accelerated for 6 seconds over a distance of

tatuchka [14]

Answer:

Explanation:

4.4 x 6x3=-03o4-

4 0

3 years ago

What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the

alex41 [277]

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter d = 20 cm

so, radius r = 10 cm = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

sphere volume $V = \frac{4}{3} \pi r^3$

$=\frac{4}{3} \pi 0.10^3$

=4.18*10^{-3} m^3

we know that

$Density = \frac{Mass}{Volume}$

mass M = Density * Volume

= (300)(4.18*10^{-3} m3)

=1.25 kg

mass of the water displace = volume *density of water

= 4.18*10^{-3} m3 * 1000

= 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg -1.25 kg

= 2.93 kg

3 0

3 years ago

A speedboat initially at rest accelerates uniformly at 4.0 m/s (squared) for 7.0 s. How fast is the boat moving after 7.0 s?

zaharov [31]

Well if the boat initially at rest accelerates at uniformly at 4.0 m/s (squared) then it would be best to muitlply it so 4.0 squared equals 2 by multiplying that by 7.0 your answer would be 14 s

8 0

3 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

$\sf \implies s = 100 \: m$

6 0

3 years ago

A lizard accelerates from 2 m/s to 10 m/s in 4 seconds. what is the lizard average acceleration

VLD [36.1K]

Acceleration = (change in speed) / (time for the change)

change in speed = (ending speed) - (starting speed)

change in speed = (10 m/s) - (2 m/s) = 8 m/s

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s²)

<em>Acceleration = 2 m/s²</em>

8 0

3 years ago