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Montano1993 [528]
3 years ago
10

PLEASE HELP! I'LL GIVE BRAINLEST​

Physics
1 answer:
Tema [17]3 years ago
4 0

Answer:

a - is the amount of matter in this object

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Whereas the vast majority of astrophysicists support the big bang theory, many nonscientists consider the theory to be controver
lubasha [3.4K]

Answer:

Religion

Explanation:

Religion is a big factor. Many people believe that the Big Bang DID happen while some believe it to be incorrect due to religious beliefs. I can guarantee if you ask people on the street about the Big Bang, someone will say something about religion. The Theory goes against every religions beliefs as to how the universe was created.

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3 years ago
Which of the following are examples of projectile motion?
denis-greek [22]

Answer:

A or C    I would pick C

Explanation:

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4 years ago
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The tension at which a fishing line snaps is commonly called the line's strength .what minimum strength is needed for a line tha
marusya05 [52]
Equal the tension of the line to the force needed to stop (deselerate) the mass from a speed of 9.2 ft/s to 0 in 4.4 inches.

Start converting all the information to the SI.

mass, m = 19 lb * [1kg/2.2046lb] = 8.618 kg

speed: 9.2 ft/s * .3048m/ft = .2.80 m/s

distance to stop: 4.4 in * [2.54cm/in]*[1m/100cm] = 0.11176 m

constant acceleration => Vf^2 = Vo^2 - 2ad

Vf = 0 => a= V0^2 / 2d = [2.80m/s]^2 / (2*0.11176m) = 35.1 m/s^2

Use the second Law of Newton

Net force = m*a = 8.618 kg * 35.1 m/s^2 = 302.5 N

Answer: 302.5 N

 
3 0
3 years ago
Suppose you apply a flame and heat one liter of water, raising its temperature 10°C. If you transfer the same heat energy to two
polet [3.4K]
(a) the water density is d=1000 kg/m^3, and 1 liter corresponds to a volume of V=1 L=0.001 m^3. Therefore we can find the mass of the water in the first case:
m=dV=(1000 kg/m^3)(0.001 m^3)=1 kg

The amount of heat supplied to the water to raise its temperature by \Delta T=10 ^{\circ} C is
Q=m C_s \Delta T (1)
where 
C_s = 4.18 kJ/(kg ^{\circ} C} is the specific heat capacity of the water. 
Using the data, we find
Q=(1 kg)(4.18 kJ/(kg ^{\circ} C)(10^{\circ} C)=41.8 kJ

We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
m=dV=(1000 kg/m^3)(0.002 m^3)=2 kg
And so by re-arranging equation (1) we can calculate the new increase of temperature:
\Delta T_2 =  \frac{Q}{m C_s}  =  \frac{41.8 kJ}{(2 kg)(4.18 kJ/(kg ^{\circ} C)}=5 ^{\circ} C

(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
m=dV=(1000 kg/m^3)(0.003 m^3)=3 kg
And so, the increase in temperature if we use the same amount of heat as before is
\Delta T_3= \frac{Q}{m C_s} = \frac{41.8 kJ}{(3 kg)(4.18 kJ/(kg ^{\circ} C)}=3.3 ^{\circ} C
5 0
4 years ago
A 54 kg pig runs at a speed of 1.0
olchik [2.2K]

Answer:

27 Joules.

Explanation:

use the formula for kinetic energy:

KE = 1/2mv^2

3 0
4 years ago
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