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liubo4ka [24]
3 years ago
14

Jessica has a mass of 55kg she sleds down a hill that has a slope of 32 degrees. what is the component of her weight that is alo

ng her direction of motion ?

Physics
2 answers:
Snowcat [4.5K]3 years ago
8 0

Answer:

286 rounded, - A P E X

Explanation:

Arturiano [62]3 years ago
7 0

Answer:

W = 285.62 N

Explanation:

It is given that,

Mass of Jessica is 55 kg

Slope of the hill is 32 degrees

We need to find the component of her weight that is along her direction of motion.

The component along her direction of motion is shown in attached figure. It means

W_y=mg\sin\theta\\\\W_y=55\times 9.8\times \sin(32)\\\\W_y=285.62\ N

So, the component of her weight that is along her direction of motion is 285.62 N.

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Which of the following cart configurations could produce the X versus T graph shown
Xelga [282]

Answer:

B. Cart B

Explanation:

8 0
3 years ago
Ball A (mass of 0.5 kg) going +2.0 m/s collides with a stationary ball B (mass of 0.4 kg).
ki77a [65]

The correct answer is 1.2 m/s

: mv+mv=mv+mv

(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)

= 1kg*m/s=(0.5kg)v+0.4kg*m/s

=1kg*m/s-0.4kg*m/s=(0.5kg)v

=0.6kg*m/s=(0.5kg)v

to solve for v we divide both side by 0.5kg

v=1.2m/s

6 0
3 years ago
How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?
alekssr [168]

Answer:

1.2646\times10^{25}\ atoms

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is Mo(Cr_2O_7)_3

-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate

-We apply Avogadro's constant to find the number of atoms of oxygen:

Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms

Hence, there are 1.2646\times10^{25} \ atoms

3 0
3 years ago
a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

Heat Supplied Q=10 J

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

thus \Delta U=mc_v\Delta T=Q

c_v for mono-atomic gas is 12.471 J/K-mol

n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

3 0
3 years ago
By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged
Nutka1998 [239]
<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then: 

F = G(Mm / (4(pi)*r^2)) 

The above expression gives the force that you feel on the earth's surface, as it is today! 

Let us now double the mass of the earth and decrease its diameter to half its original size. 

This is the same as replacing M with 2M and r with r/2. 

Now the gravitational force (F' ) on the new earth's surface is given by: 

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F 

So: 

F' = 8F 

This implies that the force that you would feel pulling you down (your weight) would increase by 800%! 

You would be 8 times heavier on this "new" earth!</span>
4 0
3 years ago
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