Answer:
Explanation:
According to the principle of energy conservation
all kinetic energy will change into thermal energy to increase its temperature
Next, divide both sides by the object's mass
the resulting temperature change is expressed as
An astronaut is standing on the surface of a planet that has a mass of 6.42×1023 kg and a radius of 3397 km. The astronaut fires
1 answer:
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Answer:
Speeds of 1.83 m/s and 6.83 m/s
Explanation:
Based on the law of conservation of momentum,
where m represents mass,
is the initial speed before impact,
and
are the velocities of the impacted object after the collision and of the originally stationary object after the impact.
Thus,
After the collision, the kinetic energy doubles, therefore:
Substituting the initial velocity of 5 m/s provides the equation needed to proceed.
We know that
leads to
Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25.
By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.
Answer:
10000 V
0.00225988700565 m²
Explanation:
E = Electric field =
d = Distance = 2.5 mm
Q = Charge = 80 nC
= Permittivity of free space =
The potential difference is calculated as
The potential difference across the plates amounts to 10000 V
Area is determined by
The area of each plate measures 0.00225988700565 m²
Capacitance is determined by
The capacitance is
Answer:
130 m/s (to two significant figures)
Explanation:
In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.
u represents the initial projectile velocity = 150 m/s
uₓ = u cos θ = 150 cos 30° = 129.9 m/s
uᵧ = u sin θ = 150 sin 30° = 75.0 m/s
A projectile's motion can be viewed as made up of independent vertical and horizontal elements.
The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.
Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.
Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.
At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s
