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Nady [450]
3 years ago
13

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resistin

g motion, including friction and air resistance, total 432.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 241 kg
Physics
1 answer:
otez555 [7]3 years ago
7 0

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

a=3 \ m/s^2

Forces resisting motion,

F_f=432 \ N

Mass,

m = 241 \ kg

By using Newton's second law, we get

⇒ F-F_f=ma

Or,

⇒         F=ma+F_f

By putting the values, we get

⇒             =(3\times 241)+432

⇒             =723+432

⇒             =1155 \ N

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son4ous [18]

Answer:

The answer is "36 grams".

Explanation:

In this question, the weight of the ball is not mentioned but is the weight of the cookies is declared, which is equal to 36 grams, and all the cookies are squeezes into the ball and after that, it calculates the overall weight so, let assume that ball weight is =0 and then the overall weight is:

=\text{weight of ball + cookies weight}\\\\=0+36 \ grams \\\\=36 \ grams

5 0
3 years ago
2. A ball tied to a pole by a rope swings in a circular path with a centripetal acceleration of 2.7 m/s. If the ball has a
Helga [31]

Answer: The diameter of the circular path is 2.96m

Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.

Centripetal acceleration = 2.7m/s^2

Tangential speed = 2.0m/s

Radius = 2.0^2 / 2.7 = 4/2.7

= 1.48m

Diameter = radius*2

= 1.48*2 = 2.96m.

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3 years ago
Which statements did both Aristotle and Ptolemy assume? Check all that apply.
tensa zangetsu [6.8K]

Bodies in space traveled in circles.

The planets revolved around the Earth.

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3 years ago
Read 2 more answers
How fast (in rpm) must a centrifuge rotate if a particle 7.3 cm from the axis of rotation is to experience an acceleration of 1.
guajiro [1.7K]

The formula we use here is:

radial acceleration = ω^2 * R <span>

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m 
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>

<span>and since: ω = 2pi*f  --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm 
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3 years ago
A 3.20 kg block starts at rest and slides a distance d down a frictionless β = 30.0 ◦ incline, where it runs into a spring with
Virty [35]

Answer:

Explanation:

a ) work done by gravitational force

= mg sinθ ( d + .21)

Potential energy stored in compressed spring

= 1/2 k x²

= .5 x 431 x ( .21 )²

= 9.5

According to conservation of energy

mg sinθ ( d + .21)  = 9.5

3.2 x 9.8 x sin 30( d + .21 ) = 9.5

d = 40 cm

b )

As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.

mg sin30 = kx

3.2 x 9.8 x .5 = 431 x

x =  3.63 cm

When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.

7 0
3 years ago
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