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3 years ago

The gravitational force between two objects increases as mass

2 answers:

7 0

F=g(m1*m2)/r^2
gravitational force is directly proportional to product of masses and inversely proportional to square of distance between them.so consequently mass should increase and distance should decrease

Bezzdna [24]3 years ago

6 0

The gravitational force between two objects increases as mass increases or distance decreases. This will make the correct answer C.

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A 0.40-kg mass, attached to the end of a 0.75-m string, is whirled around in a circular horizontal path. If the maximum tension

hichkok12 [17]

To solve this problem we will apply the concepts given from the circular movement of the bodies for which we have that the centripetal Force is defined as a product between the mass and the velocity squared at the rate of rotation, mathematically this is

$F_c = \frac{mv^2}{r}$

Where,

m = Mass

v = Velocity

r = Radius

Our values are given as

$m = 0.4kg\\r = 0.75m\\F_c = 450N$

Rearranging to find the velocity we have that,

$F_c = \frac{mv^2}{r}$

$v = \sqrt{\frac{F_c * r}{m}}$

$v = \sqrt{\frac{450 * 0.75}{0.4}}$

$v = 29.0474m/s$

Therefore the maximum speed can the mass have if the string is not to break is 29m/s

3 0

3 years ago

Hello!

Mnenie [13.5K]

Answer:

$T_0=80695.17162...$

Explanation:

Given equation:

$\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124$

To solve the given equation:

$\textsf{Apply log rules}: \quad e^{\ln (x)}=x$

$\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}$

Multiply both sides by T₀:

$\implies T_0-100=T_0e^{-0.00124}$

Add 100 to both sides:

$\implies T_0=T_0e^{-0.00124}+100$

Subtract $T_0e^{-0.00124}$ from both sides:

$\implies T_0-T_0e^{-0.00124}=100$

Factor out the common term T₀:

$\implies T_0(1-e^{-0.00124})=100$

Divide both sides by $(1-e^{-0.00124})$

$\implies T_0=\dfrac{100}{1-e^{-0.00124}}$

Carry out the calculation:

$\implies T_0=\dfrac{100}{1-0.99876...}$

$\implies T_0=\dfrac{100}{0.001239231...}$

$\implies T_0=80695.17162...$

6 0

3 years ago

A 0.60-kilogram softball initially at rest is hit with a bat. The ball is in contact with the bat for 0.20 second and leaves the

Furkat [3]

Answer:

75 Newtons.

Explanation:

From Newton's second law of motion,

F = m(v-u)/t................... Equation 1

Where F = force exerted by the ball on the bat, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time

Given: m = 0.6 kilogram, u = 0 meter per seconds (at rest), v = 25 meters per seconds, t = 0.2 seconds.

Substitute into equation 1

F = 0.6(25-0)/0.2

F = 3(25)

F = 75 Newton.

Hence the magnitude of the average force exerted by the ball on the bat = 75 Newtons.

8 0

4 years ago

The scientific method is

pashok25 [27]

The conclusion of a scientific process is a statement of whether the original hypothesis was supported or refuted by the observations gathered.

The six steps are:
1) Make an observation
2) Ask a question
3) Form a hypothesis
4) Conduct an experiment
5) Accept or Reject the hypothesis
6) Share your results

7 0

4 years ago

11. Highlight your answer: In a double slit experiment with slits 1.00x10^-5m apart, light casts the first bright

grin007 [14]

Answer:

λ ≈ 462 nm

Explanation:

Given:-

- The slit separation, a = 10^-5 m

- The first bright fringe occurs at, y = 0.03 m from central order

- The distance between the screen and slit, L = 0.65 m

Find:-

what is the wavelength of this light?

Solution:-

- The Young's double slit experiment gives us the relation for the interaction of a light with wavelength ( λ ) that pass through 2-slits with separation ( a ) and interfere constructively or destructively at the the screen Hence, forming an interference pattern of bright and dark fringes.

- The relation for separation ( y ) between bright fringes is related to the above mentioned parameters as given below:

y = L*n*λ / a

Where,

n: The order of fringe from central

Re-arrange for wavelength ( λ ).

- The order of first bright fringe is n = 1:

λ = y*a / L

λ = 0.03*0.00001 / 0.65

λ = 4.61538 * 10^-7 m

λ ≈ 462 nm

6 0

4 years ago