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Ber [7]
3 years ago
5

PLS HELP ME!!!

Physics
1 answer:
natka813 [3]3 years ago
3 0
Hmmmmmmmmm i don’t know
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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
The mass of the Moon is 7.35 x 1022 kg, while that of Earth is 5.98 x 1024 kg. The average distance from the center of the Moon
Kryger [21]

Answer:

aaa

Explanation:

m_e = Mass of the Earth =  5.98 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r_1 = Distance from the center of the Moon to the center of Earth = 6371000 m

r_2 = Distance from the center of the earth center to sun center

m_m = Mass of moon = 7.35\times 10^{22}\ kg

M = Mass of sun = 1.989\times 10^{30}\ kg

F_1=G\frac{m_em_m}{r_1^2}\\\Rightarrow F_1=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 7.35\times 10^{22}}{(384000000)^2}\\\Rightarrow F_1=1.988\times 10^{20}\ N

F_2=G\frac{Mm_e}{r_1^2}\\\Rightarrow F_2=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 1.989\times 10^{30}}{(149.6\times 10^9+6371000+695.51\times 10^6)^2}\\\Rightarrow F_2=3.511\times 10^{22} N

\frac{F_1}{F_2}=\frac{1.988\times 10^{20}}{3.511\times 10^{22}}\\\Rightarrow \frac{F_1}{F_2}=0.00566\\\Rightarrow F_1=F_20.00566

Hence the force of moon on earth is 0.00566 times the force of earth on moon center to center

4 0
4 years ago
A 3 kg ball rolls at 2m/s does this system favor kinetic energy or potential energy?
MissTica

Answer:

Kinetic energy

Explanation:

Total energy of any  system is always conserved.

E = K +U

E= Total energy

K = kinetic energy

U = Potential energy

Kinetic energy is the energy acquired by the system due to the virtue of it's motion.

Potential energy is the energy acquired by the system due to the virtue of it's configuration.

Also, Kinetic energy K = \frac{1}{2} m v^{2}

K = \frac{1}{2} \times 3 \times 2^{2}

K = 6 J

5 0
3 years ago
A person walks 100m in 5
RUDIKE [14]
Answer:
20.35m per minute
Or
1,221.43 per hour





100/5=20
200/7= 28.57
50/4= 12.5

20+28.57+12.5= 61.07

61.07/3= 20.35m per minute

Or

20.35 x 60= 1,221.43m per hour
6 0
3 years ago
How much heat is absorbed by 60 g of copper when it is heated from 20°C to 80°C?
ira [324]

Answer:

1,836J

Explanation:

6 0
4 years ago
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