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3 years ago

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HELLO .................. OVER HERE ................ HELP NEEDED ............... The mass of 7.5 kg has weight of 30 N on a certa

in planet. Calculate the acceleration due to gravity on this planet.

1 answer:

sveticcg [70]3 years ago

3 0

Answer:

4 kg

Explanation:

Equation for kg to N

7.5 * x = 30

Solve for x

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The specific heat of acetic acid is 2.07 J/gºC. If 1150 J of heat is

Doss [256]

Answer:

12.3 °C

Explanation:

Q=m*Cp*ΔT-->

1150=45*2.07*ΔT-->

ΔT=1150/93.15-->

ΔΤ=12.3 °C

4 0

3 years ago

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2

Shtirlitz [24]

99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
= (0²-27.5²)/(2x50.0)
=-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!

4 0

3 years ago

A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m 2, point, 0, start frac

dem82 [27]

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

7 0

3 years ago

Read 2 more answers

Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0

3 years ago

A brick of mass 2.0kg is at rest. It falls to the ground through a

Lisa [10]

Answer:

I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

I hope my little bit (big you may say) hint help you with your question.

5 0

3 years ago