The volume of the object must be no larger than 
.
Explanation:
In order for an object to be able to float in water, its density must be equal or smaller than the water density.
The density of water is:
This means that the density of the object must be no larger than this value.
We also know that the density of an object is given by
where
m is the mass of the object
V is its volume
For the object in this problem, the mass is
Therefore, we can re-arrange the equation to find its volume:
So, the volume of the object must be no larger than .
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The speed of a proton after it accelerates from rest through a potential difference of 350 V is 
.
Initial velocity of the proton 
Given potential difference 
let's assume that the speed of the proton is 
,
Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge 
when accelerated with a potential difference 
is,
Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e
If the initial and final velocity of the proton is 
and respectively then,
change in Kinetic Energy 
change in Potential Energy 
from conservation of energy,
so, 
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Answer:
<h2>
14.66secs</h2>
Explanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
The refrigerator's coefficient of performance is 6.
The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.
As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.
The heat produced in the cold compartment, H = 780.0 J
Work done in ideal refrigerator, W = 130.0 J
Refrigerator's coefficient of performance = H/W
= 780/130
= 6
Therefore, the refrigerator's coefficient of performance is 6.
Energy conservation requires the exhaust heat to be = 780 + 130
= 910 J
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