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Alchen [17]
3 years ago
11

The mass of the bicycle and rider is 60 kg and the total area of the tyres in contact with

Physics
1 answer:
AleksAgata [21]3 years ago
3 0

Answer:

a little

Explanation:

First of all, it's not how you spell "tyres", it is tires.

Second of all, you already know the Mass so what you need to find out now is  contact the road. You Know that your number and letter are squared so that would turn into 6m x 2.4. Then you do the math do continue on to finish it. Have a great day!! Good luck with the answer!!

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A bus travels north on some busy city streets for 2.5 km, and a trip
d1i1m1o1n [39]

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

Hence, the velocity of the  bus, V = 11.31 m/s

8 0
3 years ago
12)
Nataliya [291]

Answer: b

Explanation: the two pieces will repel as both have obtained a static charge.

5 0
3 years ago
What if the Earth was moved out to Jupiter's orbit which is about 5
Stels [109]

Answer:

25.0 less force

Explanation:

4 0
3 years ago
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N
guajiro [1.7K]

Answer:

\frac{r_1}{r_2}=6.9

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9

4 0
3 years ago
Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f
Step2247 [10]
The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.
8 0
4 years ago
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