What happens when calcium reacts with chlorine?

When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the ball was in contact wi

SVEN [57.7K]

when the ball hits the floor and bounces back the momentum of the ball changes.

the rate of change of momentum is the force exerted by the floor on it.

the equation for the force exerted is

f = rate of change of momentum

$f = \frac{mv - mu}{t}$

v is the final velocity which is - 3.85 m/s

u is initial velocity - 4.23 m/s

m = 0.622 kg

time is the impact time of the ball in contact with the floor - 0.0266 s

substituting the values

$f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}$

since the ball is going down, we take that as negative and ball going upwards as positive.

f = 189 N

the force exerted from the floor is 189 N

4 0

3 years ago

A is a hot object. B is a cold object. the two are brought into contact with each other so that heat is developed

Umnica [9.8K]

Answer:

A) Cold object will start getting hot

B) Heat exchange will stop as the two object acquire the same temperature.

Explanation:

A) When one hot object and one cold object are kept in contact then the heat is transferred from the hot object to the cold object via different modes of heat transmission. Hence, the cold object starts getting hot

B) The transmission of heat from the hot object to the cold object will stop as the temperature of the two object becomes equal to each other.

3 0

3 years ago

If one object is 103 km away and a second object is 106 km away, one could say that the second object is _____ times further awa

vaieri [72.5K]

Answer:

$1.03$

Explanation:

$\frac{object_{second}}{object_{first}} = \frac{106}{103} = 1.02912621359$

Round to three significant digits

$1.03$

7 0

3 years ago

A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr

jasenka [17]

Answer: The original temperature was

$T_{1}=126.51K$

Explanation:

Let's put the information in mathematical form:

$V_{1}=12.7m^{3}$

$T_{1}=?$

$V_{2}=32.5m^{3}$

$T_{2}=323K$

$P_{1}=P_{2}=3atm$

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

$PV=nRT$

were R is the gas constant. And n is the number of moles (which we don't know yet)

From this, taking $R=0.08205746\frac{atm.l}{mol.K}$ , we have:

$n=\frac{P_{2}V_{2}}{RT_{2}}$

⇒ $n=3.67mol$

Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒ $T_{1}=126.51K$

7 0

3 years ago

Read 2 more answers

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

Using the Gaussian equation, to know where the object is situated (distance from the point).

$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago