Which of the following planets orbits the fastest around the sun?

3 3

olganol [36]

Answer:

I Dont know u answer it

Explanation:

8 0

2 years ago

Which formula can be used to solve problems related to the first law of thermodynamics?

docker41 [41]

Answer: The formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

$\Delta U=Q+W$

where, Q = heat energy

W = work energy

$\Delta U$ = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For $\Delta U$ : When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

8 0

3 years ago

Read 2 more answers

The uniform 100-kg I-beam is supported initially by its end rollers on the horizontal surface at A and B. by means of the cable

Ann [662]

I attached a free body diagram for a better understanding of this problem.

We start making summation of Moments in A,

$\sum M_A = 0$

$P(6cos\theta)-981(4cos\theta)=0$

$P=654N$

Then we make a summation of Forces in Y,

$\sum F_y = 0$

$654+R-981 = 0$

$R=327N$

At the end we calculate the angle with the sin.

$sin\theta = \frac{3m}{4m+2m+2m} = \frac{3m}{8m}$

$\theta = 22.02\°$

8 0

3 years ago

A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m

slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 $m/s^{2}$ and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

$M\frac{dv}{dt} = u\frac{dM}{dt}$

$\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt$

= $u\int_{M_{o}}^{M_{f}} \frac{dM}{M}$

$v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}$

$v_{o} = 0$

As, $v_{f} = u ln (\frac{M_{f}}{M_{o}})$

u = -2900 m/s

$M_{f} = M_{o} - m \times t_{f}$

= $2500 kg + 1000 kg - 95 kg \times t_{f}s$

= $(3500 - 95t_{f})s$

$v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s$

Also, we know that

a = $\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}$

= $\frac{u}{3500 - 95 t} \times (-95) m/s^{2}$

= $\frac{95 \times 2900}{3500 - 95t} m/s^{2}$

At t = 10 sec,

$v_{f}$ = 918.34 m/s

and, a = 108 $m/s^{2}$

3 0

3 years ago

A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that othe

Marina86 [1]

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

$\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9 = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C$

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

6 0

2 years ago