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lianna [129]
3 years ago
15

Why is copper a pure substance? A. Because it is an element B. Because it is a compound C. Because it is a homogenous mixture​

Physics
1 answer:
expeople1 [14]3 years ago
3 0
A) because it is an element.
Explanation: usually when they say pure substance they mean pure element.
Hope this helped!! :)
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Can someone pleaseeee answer this !!!!!!
LenaWriter [7]

Answer:

The person with locked legs will experience greater impact force.

Explanation:

Let the two persons be of nearly equal mass (say m)

The final velocity of an object (person) dropped from a height H (here 2 meters) is given by,

v=\sqrt{2gH}

(g = acceleration due to gravity)

which can be derived from Newton's equation of motion,

v^2=u^2+2aS

Now, the time taken (say t ) for the momentum ( mv ) to change to zero will be more in the case of the person who bends his legs on impact than who keeps his legs locked.

We know that,

Force=\frac{\Delta(mv)}{t}

Naturally, the person who bends his legs will experience lesser force since t is larger.

3 0
3 years ago
ANSWER THE FOLLOWING QUESTION AND MAKE SURE TO GIVE A FULL DESCRIPTION TO HOW YOU GOT YOUR ANSWER (EX. make sure to use blank ru
vazorg [7]
Here is how I did it
6 0
4 years ago
If you had a rock with a volume of 237 ml and a density of 4.52 g/ml how much mass would it have?
larisa [96]

Explanation:

Mass = Volume * Density

= 237ml * (4.52g/ml)

= 1071.2g.

7 0
3 years ago
Read 2 more answers
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
Your cousin is moving into an apartment in San Francisco. This apartment is on a street that is angled at 25∘ above the horizont
Vesnalui [34]

Answer:

70

Explanation:

5 0
3 years ago
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