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madam [21]
4 years ago
7

If an object is at rest would there be any forces acting upon it?

Physics
2 answers:
Marat540 [252]4 years ago
7 0
Yes gravity would be the force that is holding that object down
ValentinkaMS [17]4 years ago
3 0
Well, it seems as though there would be no forces, because it's "at rest", but wouldn't gravity still be acting upon it?
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Can someone help me with this Physics question please?
strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

N=N_0(\frac{1}{2})^{\frac{t}{H} where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

N=48.0(\frac{1}{2})^{\frac{49.2}{12.3} and simplifying a bit:

N=48.0(.5)^4 and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

6 0
3 years ago
Read 2 more answers
Critical angle of glass is 42 .what does it mean?​
elena55 [62]

Answer:

i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!

Explanation:

HOPE IT HELPS!!

6 0
3 years ago
As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period
Hunter-Best [27]
Periodic time is the time taken for one complete oscillation by a body in circular motion. In this case the merry-go round takes 2 minutes to cover 15 complete oscillations. 2 Minutes = 120 seconds
Hence, 15 oscillations takes 120 secs
         thus 1 oscillation takes 120/15 = 8 seconds
therefore the period of the merry-go-round = 8 seconds
8 0
3 years ago
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An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad
Anna11 [10]

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0
4 years ago
The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha
denis-greek [22]

Answer:

H=41.3kJmol^{-1}

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}

if we insert values, we arrive at

ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}

4 0
3 years ago
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