Answer:
-16°C
Explanation:
PV = nRT
V and n are constant.
P / T = P / T
(2 atm + 1 atm) / (266 K) = (1.9 atm + 1 atm) / T
T = 257.1 K
T = -16°C
Answer:
the neutron star becomes a black hole
Answer:
The work done by the weightlifter, W = 700 J
The power of the weightlifter, P = 350 watts
Explanation:
A weightlifter lifts a set of weights a vertical distance, s = 2 m
The force exerted to lift the weight, F = 350 N
The work done by the body is defined as the product of the force applied by the body to the displacement it caused.
W = F x s
= 350 N x 2 m
= 700 J
The work done by the weightlifter, W = 700 J
The time taken by the weightlifter to lift the weight, t = 2 s
The power is defined as the rate of body to do work. It is given by the equation,
P = W / t
= 700 J / 2 s
= 350 watts
Hence, the power of the weightlifter, P = 350 watts
Answer:
Minimum time interval (t2)=0.90 SECONDS
Explanation:
- coefficient of friction for employees footwear = 0.5
- coefficient of friction for typical athletic shoe = 0.810
- frictional force = coefficient of friction X acceleration due to gravity X mass of body
- Acceleration due to gravity is a constant = 9.81 m/s
- Let frictional force for employee footwear = FF1
- Let frictional force for athletic footwear =FF2
FF1 = O.5 X 9.81 X mass of body
= 4.905 x mass of body
FF2 = 0.810 X 9.81 X mass of body
= 7.9461 x mass of body
The body started from rest there by making the initial velocity zero ( u = 0)
From d= ut + 1/2 a x 
- d =
x a x
.....................................i
where d= distance and it is given as 3.25m
- F =ma ...................................ii
making acceleration subject of the formula from equation ii
- a =

Making t subject of formula from equation (i)
- t=

where
-
= 4.905
=7.9461
Let
- t1 = minimum time taken for frictional force for employee foot wear
- t1 =
=1.15 seconds
- t2 =
= 0.90 seconds
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