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faltersainse [42]
3 years ago
7

Who’s of the following is classified as a salt: ammonia or sodium chloride?

Physics
1 answer:
Temka [501]3 years ago
8 0

Sodium Chloride could be classified as a salt.

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Name an element in the same family as sodium
AlladinOne [14]

Answer:

The Alkali Metals

sodium is a member of the alkali metal family with potassium (K) and lithium (Li). Sodium's big claim to fame is that it's one of two elements in your table salt. When bonded to chlorine (Cl), the two elements make sodium chloride (NaCl).

7 0
2 years ago
Read 2 more answers
A 24 kg child descends a 5.0 m high slide and reaches the ground with a speed of 2.8 m/s. What is the mass of the child?
laila [671]
The mass of the child in this problem is 24 Kg, the unit for mass is in Kilogram.
The mass will change only if it travels close to the speed of light.
7 0
3 years ago
James Bond is trying to escape his enemy on a speedboat but
SVEN [57.7K]

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is  completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Putting all the values, we get :

0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s

So, Bond's boat is moving with a speed of 100 m/s after the collision.

3 0
3 years ago
Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o
VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so,  we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

K=Celsius+273.15

So, converting we have:

T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K

Therefore, substituting the given information and calculating, we have:

HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )

HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0
3 years ago
What is the momentum of a 15 kg tire rolling down a hill at 3 m/s? 5 kg • m/s 18 kg • m/s 45 kg • m/s 60 kg • m/s
maw [93]

p=mv

?=15kg * 3m/s

          45p

7 0
3 years ago
Read 2 more answers
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