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Sav [38]
3 years ago
14

With what magnitude of force does a ball of mass 0.75 kilograms need to be hit so that it accelerates at the rate of 25 meters/s

econd2? Assume that the ball undergoes motion along a straight line.
Physics
2 answers:
SSSSS [86.1K]3 years ago
8 0

Answer:

Assume that the ball undergoes motion along a straight line. ... F = m A Force = (mass) x (acceleration) The question tells you the mass and the acceleration. All YOU have to do is take the numbers and pluggum into Newton's 2nd law. F = m A = (0.75 kg) (25 m/s²) = (0.75 x 25) kg-m/s² = 18.75 Newtons .

Explanation:

i looked it up ok

Zielflug [23.3K]3 years ago
8 0

<u>Answer: </u>

A ball of mass 0.75 kilograms needs to be hit so that it accelerates at the rate of 25 meters/second^2 assuming that the ball undergoes motion along a straight line and the mount of Force required is 18.75 newton.

<u>Explanation:</u>

According to Newton`s 2nd  law F=Ma,

where M mass of the object, a- Acceleration due to gravity.

Given-

Mass (m) = 0.75 kg

Acceleration (a) = 25 m/s^2

We know That, Force = mass × acceleration

Substituting the given values , we get

  F = 0.75 × 25

   F= 18.75 N  

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The maximum velocity in a banked road, ignoring friction, is given by;

v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.

Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°

Therefore, the road has been banked at 5.24°.
4 0
4 years ago
Carol has several test tubes of different liquids. If Carol removes heat from the liquid substances, which of the following is m
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Answer:

d

Explanation:

6 0
3 years ago
The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopr
OlgaM077 [116]

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

(e) displacement amplitude of the initial note 4.78*10^(-7) meters

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valentina_108 [34]
Three 40w lamp for 6 hours
8 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
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