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Damm [24]
2 years ago
15

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

rtunately, you have the density of mercury in units of kilogram-meter^3 and the density of silicon in other units: 2.33 gram-centimeter^3. You decide to convert the density of silicon into units of kilogram-meter^3 to perform the comparison.
By which combination of conversion factors will you multiply 2.33 gram-centimeter^3 to perform the unit conversion?
Physics
1 answer:
dolphi86 [110]2 years ago
3 0

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

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A particle of mass M is moving in the positive x direction with speed v. It spontaneously decays into 2 photons, with the origin
anygoal [31]

Solution :

Mass of the particle = M

Speed of travel = v

Energy of one photon after the decay which moves in the positive x direction = 233 MeV

Energy of second photon after the decay which moves in the negative x direction = 21 MeV

Therefore, the total energy after the decay is = 233 + 21

                                                                           = 254 MeV

So by the law of conservation of energy, we have :

Total energy before the decay = total energy after decay

So, the total relativistic energy of the particle before its decay = 254 MeV  

7 0
2 years ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
2 years ago
1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the
Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0
2 years ago
Every spring has an equilibrium position. Which statements describe a spring at its equilibrium position? Check all that
alexgriva [62]

Answer:

The elastic potential energy is zero.

The net force acting on the spring is zero.

Explanation:

The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.

Let's now analyze the different statements:

The spring constant is zero.  --> false. The spring constant is never zero.

The elastic potential energy is at a maximum  --> false. The elastic potential energy of a spring is given by

E=\frac{1}{2}kx^2

where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.

The elastic potential energy is zero.  --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).

The displacement of the spring is at a maxi num  --> false, for what we said above

The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched

8 0
2 years ago
Read 2 more answers
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
2 years ago
Read 2 more answers
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