Answer:
= 15.57 N
= 2.60 N
= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Explanation:
The weight of the sugar bag on Earth is:
g=9.81 m/s²
m=3.50 lb=1.59 kg
=m·g=1.59 kg×9.81 m/s²= 15.57 N
The weight of the sugar bag on the Moon is:
g=9.81 m/s²÷6= 1.635 m/s²
=m·g=1.59 kg× 1.635 m/s²= 2.60 N
The weight of the sugar bag on the Uranus is:
g=9.81 m/s²×1.09=10.69 m/s²
=m·g=1.59 kg×10.69 m/s²= 16.98 N
The mass of the bag is the same on the three planets. m=1.59 kg
Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:
Explanation:
Set up force equation
∑F=ma
∑F=W+FB
The minus sign for downward direction
Answer:
(1) 10^−2 m
Explanation:
The diameter of the tire of an automobile is generally expressed in centimetres; we can say that the diameter of a tire is generally about
d = 20 cm (20 centimetres)
Now we have to verify which option is closest to this value. To do that, we have to keep in mind the equivalence between metres and centimetres; in fact, we have:
This means that we can rewrite the diameter of the tire of a car as
By comparing it with the given options, we see that the closest option is
(1) 10^−2 m
which is therefore the correct answer.