Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
Answer: E= KQ/r^2
Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.
The magnitude of an electric field E, at a point is given by Coulombs law as
E/ F/q
Where F= Coulombs force exertedon the charge and q= electric charge
E= F/q=(KQq)/r^2q
E=KQ/r^2
Answer: B. The gravitational field strength of Planet X is Wx/m.
Explanation:
Weight is a force, and as we know by the second Newton's law:
F = m*a
Force equals mass times acceleration.
Then if the weight is:
Wx, and the mass is m, we have the equation:
Wx = m*a
Where in this case, a is the gravitational field strength.
Then, isolating a in that equation we get:
Wx/m = a
Then the correct option is:
B. The gravitational field strength of Planet X is Wx/m.
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³
Therefore:
8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.
Answer: 8.06 in³=1.321 x 10⁻⁴ m³
Answer:
4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period
Explanation:
The creosote bush depends on sunlight to produce the food they require through photosynthesis. The shade from the solar panels would reduce the amount of sunlight that the bush receives. This would increase the mortality of the bush.
In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.
Hence, the answer is 4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period.
