[tex]\huge\boxed{Option D}[/tex]
_____________________________________DATA:Radius of Mercury = [tex]R_m[/tex] = [tex]2.43x10^6m[/tex]
Mass of Mercury = [tex]M_m = 3.2x10^{23}m[/tex]
Distance Satellite above the surface of the Mercury = d = 265,000m
Gravitational Constant = [tex]G = 6.67x10^{-11} \frac{N.m^2}{kg^2}[/tex]
_____________________________________SOLUTION:Since the Satellite is orbiting around the Planet Mercury, due to the centripetal force, and Centripetal force is the force that acts towards the center of the circle, Whereas The gravitational force also acts towards the center of the circle thus we can say that Centripetal force is equal or same as centripetal force. So,
[tex]F_g =F_C[/tex]
Fg is Given by,
[tex]F_g = \frac{GM_MM_S}{r^2}[/tex]
Fc is Given by,
[tex]F_c=\frac{M_SV^2}{r}[/tex]
Where,
G is Gravitational Constant
[tex]M_e[/tex] is mass of Planet Mercury
[tex]M_S[/tex] is Mass of Satellite
r(small letter) is the distance between the center of the Planet Mercury and the satellite.
V is velocity of satellite
_____________________________________Now,
[tex]\frac{GM_MM_S}{r^2} =\frac{M_SV^2}{r}[/tex]
[tex]V = \sqrt\frac{GM_M}{r}[/tex]
r can also be written as,
[tex]V = \sqrt\frac{GM_M}{R_M +d}[/tex]
Substitute the variables,
[tex]V = \sqrt{\frac{(6.67x10^{-11})x(3.2x10^{23})}{2695000}[/tex]
Simplify the equation,
V = 2814 [tex]\frac{m}{s}[/tex]
Approximately,
V = 2800 [tex]\frac{m}{s}[/tex]
_____________________________________Best Regards,'Borz'A car of mass 1500 kg starting from rest can reach a speed of 20
m/s within 10 seconds. Calculate the accelerating force of the car
engine.
Explanation:
F=MA
F=1500 * 2
F=3000N
what is the KE of a 1.00 kg hammer swinging at 20.0 m/s? 200 Joules
I know the answer I just need help understanding it.
Answer:
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 1250 kg
Velocity (v) = 20.0 m/s.
Kinetic energy (K.E) =?
Kinetic energy is simply defined as energy possed by a body in motion. Mathematically, it is expressed as:
K.E = ½mv²
Where:
K.E is the kinetic energy
m is the mass of the object
v is the velocity of the object.
Thus, we can obtain the kinetic energy of the automobile by using the above formula as illustrated below:
Mass (m) = 1250 kg
Velocity (v) = 20.0 m/s.
Kinetic energy (K.E) =?
K.E = ½mv²
K.E = ½ × 1250 × 20²
K.E = 625 × 400
K.E = 250000 J
Therefore, the kinetic energy of the automobile is 250000 J
A substance will take on the shape of an open container if it is a
liquid.
plasma.
gas.
solid.
Answer:
liquid
Explanation:
a motorbike can travel 1000 meters in 10 minutes calculate how car it can travel in 1 sec.
Answer:
1.67meter
Explanation:
if it can travel 1000 meters in 10 minutes, 10 minutes are 600 secs (10×60), 1000÷600 is 1.67
Answer:
HI
Explanation:
A rose plant inherited two alleles for white flower petals.
Which conclusion is best supported by the given information?
Answer:
a
Explanation:
Answer:
A or each parent had at least one allele for white pedals
Explanation:
What wavelength would a ripple in water have if the frequency is 1.8 Hz and a
wave speed of 825 m/s?
Explanation:
825m/s / 1.8Hz = 458.33m
λ=v/f
λ-wavelength
v-speed
f-frequency
λ=825/1.8=458.33m
When momentum is conserved it is called _____. (multiple choice)
A.) Conservation of Momentum
B.) The Law of Momentum
C.) The Physics of Momentum
D.) The Rules of Momentum
Answer:
Based off the word "conserved" I would say
A. Conservation of Momentum.
Explanation:
Answer:
A.) Conservation of Momentum
A box is sitting stationary on a long level ramp on level ground. The coefficient of static friction is (1.0). One end of the ramp is slowly lifted higher and higher. What is the angle of the ramp with respect to the ground when the box begins slidig?
Answer:
45 degrees
Explanation:
Given that the coefficient of friction, [tex]\mu=1.0[/tex]
Let the angle of the ramp be [tex]\theta[/tex].
The gravitational force acting downward [tex]=mg[/tex]
The normal reaction by the ramp on the box, [tex]N=mg\cos\theta[/tex]
So, the maximum frictional force that can act on the box, [tex]f= \mu N[/tex]
The force along with the plane in the direction of sliding, [tex]F = mg\sin\theta[/tex]
When the box begins sliding, the F must have to overcome the frictional force,f.
So, F=f
[tex]mg\sin\theta=\mu N \\\\mg\sin\theta=\mu mg\cos\theta \\\\\frac {\sin\theta}{\cos\theta}=\mu \\\\\tan\theta=\mu \\\\\theta=\tan^{-1}\mu \\\\[/tex]
Putting the value of \mu, we have
[tex]\theta=\tan^{-1}1[/tex]
[tex]\theta=45[/tex] degrees.
Hence, the angle of the ramp with respect to the ground when the box begins sliding is 45 degrees.
the velocity of a body of mass 60kg reaches 15m/s from 0m/s in 12 second. calculate the kinetic energy and power of the body.
Answer:
KE=1/2m v^2
1/2*60*15*15
30*15*15
6750 joules
power=work/time
Jenny puts a book on her desk. Jenny’s book has an area of 200 cm2.It exerts a pressure of 0.05 N/cm2 on the desk. What is the weight of the book? *
Bob can run the 100 meter dash in 25 seconds. What is his speed?
Answer:
4 meters a second
Explanation:
100/25
plzz mark brainiest
A car travels 14 km due west, then does a
U-turn, and travels 43 km due west.
I. What total distance has the car
traveled?
II.
What is the total displacement of the
car?
III. If the entire trip took 3.20 hours.
determine the average speed of the
car. Give
your answer in both
km hour, and m s. (show
conversion)
IV. If the entire trip took 2.50 hours.
determine the average velocity of
the car. Give your answer in both
km h, and mph (show conversion)
Explanation:
It is given that,
A car travels 14 km due west, then does a U-turn and travels 43 km due east (when it takes U-turn, it will change direction from west to east)
i. Total distance = total path covered
= 14 km + 43 km
= 57 km
ii. Let east is positive and west is negative.
Displacement = final position-initial position
= 43-14
= 29 km
iii. If time taken in the entire trip = 3.2 hours
Average speed = distance/time
[tex]s=\dfrac{57\ km}{3.2\ h}\\\\=17.81\ km/h[/tex]
1 km/h = 0.2777 m/s
17.81 km/h = 4.94 m/s
iv. If time taken in the entire trip = 2.5 hours
Average velocity = displacement/time
[tex]v=\dfrac{29\ km}{2.5\ h}\\\\=11.6\ km/h[/tex]
1 km/h = 0.621 mph
11.6 km/h = 7.2 mph
Henc, this is the required solution.
What connects the colon to the anus?AColon B Anus C Rectum D Biceps
The answer is C. Rectum.
Answer:
You have a severe case.
Explanation:
My anus smells foul, like poop or diarrhea. Why? Like your anus might be tired of making the same old sensational brown. No fear needed! Anus Maker is here! With this app, you can post poop photos for help if you need! No more foul sensational browns! Start your free Anus Maker trial today!
A red 120 kg bumper car moving at 4 m/s collides with a green 100 kg bumper car moving at 3 m/s. The red bumper car bounces off at 2 m/s. What is the green car's final velocity?
plzzzz help me this is due today. only need help with these two questions and the element is oxygen.
How do protons, neutrons, and electrons differ in terms of their electrical charges and locations within the atom?
Describe the four fundamental forces. Which of these forces are involved in chemical bonding?
Answer: thanks for the point
Explanation:
deducing the acceleration = deduce the gradient of velocity-time graphs
true or false?
Answer:
True
Explanation:
Understanding the relationship between change in velocity with time will explain to you how the object is accelerating or decelerating. This means acceleration is a ratio calculated from the change in velocity of an object and change in time. In a velocity time graph, the y-axis represents the velocity and the x-axis represents the time.
The slope of the graph m = Δ y-axis values/Δ x-axis values
m= Δvelocity / Δ time
m= Δ v / Δ t ------this the definition of acceleration so;
a= Δ v / Δ t
Conclusion : The slope/gradient of a velocity -time graph is acceleration.
An example of a poor coping mechanism is ?
Answer: A bad, maladaptive, unhealthy or destructive coping mechanism is one where the behavior does not resolve the problem in the long-term and may actually increase the harm. Unhealthy coping strategies may feel like they are having the desired effect in the short term.
Explanation: Yes, it was from google...
A car of mass 1000 kg travelling at a velocity of 25 m/s collides with another car of mass 1500kg which is at rest. The two cars stick and move off together. What is the velocity of the two cars after the collision?
Answer:
The velocity of the two cars is 10 m/s after the collision.
Explanation:
Law Of Conservation Of Linear Momentum
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is
P=m.v
If we have a system of bodies, then the total momentum is the sum of them all
[tex]P=m_1v_1+m_2v_2+...+m_nv_n[/tex]
If some collision occurs, the velocities change to v' and the final momentum is:
[tex]P'=m_1v'_1+m_2v'_2+...+m_nv'_n[/tex]
In a system of two masses, the law of conservation of linear momentum takes the form:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
If both masses stick together after the collision at a common speed v', then:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v'[/tex]
The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).
Knowing both cars stick and move together after the collision, their velocity is found solving for v':
[tex]\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
[tex]\displaystyle v'=\frac{1000*25+1500*0}{1000+1500}[/tex]
[tex]\displaystyle v'=\frac{25000}{2500}[/tex]
v' = 10 m/s
The velocity of the two cars is 10 m/s after the collision.
Please find attached photograph for your answer.
Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of 8000.0 N/m. The elevator falls onto the spring while subject to a frictional force (brake) of 1000.0 N. Determine the maximum compression distance, x, of the spring.
Answer:
Maximum compression distance (x) = 2.236 m (Approx)
Explanation:
Given:
Weight of frog = 4,000 N
Height = 5 m
Constant force = 8,000 N/m
Frictional force = 1,000 N
Find:
Maximum compression distance (x)
Computation:
Using Law of conservation;
mgh = 1/2(k)(x)²
4,000(5) = 1/2(8,000)(x)²
Maximum compression distance (x) = 2.236 m (Approx)
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy
Answer:
A
Explanation:
Given that a stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy. (use 9 - 10 m/s)
O at the moment of impact
2 seconds after the stone is released after the stone has fallen 40 m
when the stone is moving at 20 m/s
At the top of the hill, the P.E = mgh
P.E = 10 × 80 × m
P.E = 800m
At the moment of impact, K.E = 1/2mv^2
K.E = 1/2 × 40^2 × m
K.E = 1/2 × 1600 × m
K.E = 800m
Since both P.E and K.E are the same, we can therefore conclude that the stone's potential energy with respect to the ground is equal to its kinetic energy at the moment of impact.
The correct answer is option A.
A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction 5° north of east for an hour.
Answer:
first one 31
second one 23
Explanation:
on edge ;))
A girl pushes a wagon at constant velocity. If the
momentum of the wagon is 50 kg*m/s at a
velocity of 2 m/s, the mass of the wagon is what
What is the work energy transfer equation?
Answer:
The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)
Answer:
The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)
Explanation:
The net work done on a particle equals the change in the particle's kinetic energy:
I need a little help with this
Answer:
truck 1 has the most velocity
Explanation:
Because it weights less which means it faster and yea
A book that weighs 5 N sits on a table. What force does the table apply to the book?
Answer:
E =F.d =[1/2]mv^2
mad = [1/2]mv^2
d= v^2/2a ,v=u+at , v^2 = [at]^2 since u=0
So d = at^2/2
F = ma= 20a= 50 , a=5/2 and t=2
so d = [5/2][2^2]/2=5
Explanation:
Every action has an equal and opposite reaction. It is an action-reaction principle. Therefore the table exerts a force of 5 N on the book in order to be in stable condition.
What is Newton's third law of motion?Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.
Therefore the table exerts a force of 5 N on the book in order to be in stable condition.
The given data in the problem is ;
W is the weight of the book sits on table = 5N
N is the normal force on the book
From the equilibrium equation ;
Weight -Normal force on the book =0
Weight =Normal force on the book
The normal force on the book =5N
Hence the table exerts a force of 5 N on the book in order to be in stable condition.
To learn more about Newton's third law of motion refer to the link;
https://brainly.com/question/1077877
What distance does a biker travel if he rides at a constant speed or 22 m/s for 45 seconds?
Answer:
it would be 990 m.
Explanation:
22 m/s x 45 seconds.
HELP ASAP Which statement is true about magnetic field lines?
A. There is no consistent pattern in the lines, B. The lines form a loop from the north pole back to the north pole and from the south pole to the south pole. C. The lines point away from the south pole of a magnet and toward the north pole. D. The lines point away from the north pole of a magnet and toward the south pole.
Answer:
i think it's c
Explanation:
but I'm not sure
Answer:
The answer is D
Explanation:
A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10s. If the airplane is putting out an average force of 5.8810x10^4 N during this time, what is the average friction force exerted on the airplane by the air?
Given :
A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10 s.
If the airplane is putting out an average force of [tex]5.8810\times 10^4 \ N[/tex].
To Find :
The average friction force exerted on the airplane by the air.
Solution :
Acceleration is given by :
[tex]a = \dfrac{162-120}{2.10}\ m/s^2\\\\a = 20 \ m/s^2[/tex]
Now, force equation is given by :
[tex]F - F_{friction} = ma\\\\F_{friction} = F-ma\\\\F_{friction} = 58810 - (2450\times 20 )\\\\F_{friction} = 9810\ N[/tex]
Therefore, frictional force exerted in the airplane by the air is 9810 N.
Charles, the 75 kg trampoline artist, lands on a trampoline with a speed of 9.0 m/s.
If the trampoline behaves like a spring with a spring constant of 52,000 N/m, what maximum distance will Charles push down the trampoline before bouncing back up? (Hint: at maximum compression, Charles is not moving.)
Answer:
The maximum distance Charles will push down the trampoline ≈ 0.342 m
Explanation:
The given parameters are;
The mass of the trampoline artist, m = 75 kg
The speed with which the trampoline artist lands, v = 9.0 m/s
The value of the spring constant of the trampoline, k = 52,000 N/m
Let x represents the maximum distance Charles will push down the trampoline
Therefore, we have;
Kinetic energy = 1/2·m·v²
The kinetic energy with which the trampoline artist lands = 1/2 × 75 × 9.0² = 3037.5
The kinetic energy with which the trampoline artist lands = 3037.5 J
The potential energy stored in a spring = 1/2·k·x² = The kinetic energy with which the trampoline artist lands
∴ 1/2 × 52,000 × x² = 3037.5
∴ x = √(3037.5/(1/2 × 52,000)) ≈ 0.342
The maximum distance Charles will push down the trampoline = x ≈ 0.342 m
Nore ordered an ice cream cone.
Answer:
okay
Explanation:
okay
Answer:
Cool
Explanation: