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3 years ago

If two balls have the same volume,

1 answer:

8 0

Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

1. Ball A will have the greater density
2. Ball C and Ball D have the same density.
3. Ball Q will have the greater density.
4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

Density = Mass/Volume.

For Case 1:

Va = Vb and Ma = 2Mb
D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
Therefore, the density of ball A,
D(a) = 2D(b).
Therefore, ball A has the greater density.

For Case 2:

Vc = 3Vd,

Vd = (1/3)Vc

Md = (1/3)Mc

D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

D(c) = D(d).

Therefore, ball C and D have the same density

For Case 3:

Vp = 2Vq and Mp = Mq
D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
Therefore, the density of ball P is half the density of ball Q
Therefore, ball Q has the greater density.

For case 4:

Mx = (1/2)My
Vx = Vy

Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

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A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0

2 years ago

Some sharks can swim at average cruising speeds of three miles per hour. If a shark swam at that average speed for seven hours,

kari74 [83]

Answer:

21 miles

Explanation:

3 miles an hour for 7 hours

Its simply 7m*3m/hr=21 miles

8 0

3 years ago

If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.

olya-2409 [2.1K]

Answer:

$F=mkv$

Explanation:

Given that

$v = v_i - kx$

We know that acceleration a given as

$a=\dfrac{dv}{dt}$

$v = v_i - kx$

$\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}$

$\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}$

We know that

$F=m\dfrac{dv}{dt}$

$F=-mk\dfrac{dx}{dt}$

$F=-mkv$

So the magnitude of force F

$F=mkv$

5 0

3 years ago

A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,

tamaranim1 [39]

Answer: a) The rate constant, k, for this reaction is $0.00516s^{-1}$

b) No $t_{\frac{1}{2}}$ does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A\rightarrow products$

Given: Order with respect to $A$ = 1

Thus rate law is:

a) $Rate=k[A]^1$

k= rate constant

$0.00250=k[0.484]^1$

$k=0.00516s^{-1}$

The rate constant, k, for this reaction is $0.00516s^{-1}$

b) Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{\frac{1}{2}}=\frac{0.69}{k}$

Thus $t_{\frac{1}{2}}$ does not depend on concentration.

8 0

3 years ago

calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30 metre per seco

Anna71 [15]

Answer:

height is 45 metres

Explanation:

KE=PE

¹/²mv² = mgh

¹/²v² = gh

but v =30m/s

(30)² / 2 =10h

450=10h

h = 45m

5 0

4 years ago