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olga55 [171]
2 years ago
12

A probe is launched from earth and lands on mars the gravitational acceleration on mars is 3.7 m/s. Which statement best compare

s the mass and weight of the probe
Physics
2 answers:
Ugo [173]2 years ago
8 0
There are no correct statements among the choices you gave us.

When the probe arrives on the surface of Mars, its mass is the same
as it was on Earth, and at every place along the way.

But its weight is   (3.7/9.8) = 37.8% of what it weighed on Earth.
Natali [406]2 years ago
5 0
The answer is B)<span>The mass of the probe remains the same, but the weight of the probe decreases</span>
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What is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?
nikklg [1K]
The Energy is Kinetic Energy.

Kinetic Energy = 1/2*mv²,  Where m is mass in kg, v is velocity in m/s

Energy is 33750 Juoles,  v = 30m/s

1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²)     Using a calculator

m = 75 kg

Mass of object is 75 kg.
5 0
2 years ago
Read 2 more answers
A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow
leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

7 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
Which changes in an electric motor will make the motor stronger? Check all that apply.
ollegr [7]
To make the motor turn faster we can:
(a) increase the current
(b) use stronger magnets
(c) push the magnets closer to the coil
(d) put an iron centre piece into the coil
(e) adding more sets of coils
8 0
2 years ago
Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿
Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.  

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0
2 years ago
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