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Kruka [31]
2 years ago
7

A rock accelerates down a hill at 7 m/s2 with a force of 30.0 N. What is the mass of the rock?

Physics
1 answer:
choli [55]2 years ago
3 0

Answer:

we know thatv

F =m×a

so, m = F/a

m =30/7

m=30/7kg

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Can someone tell me what my displacement is?
natulia [17]
60017 is or answer ok you got it
3 0
3 years ago
A certain bridge is 4,224 feet long. What constant rate, in miles per hour, must be maintained in order to walk across the bridg
Archy [21]

Answer:

4miles/hour

Explanation:

the solution for this question requires that the quantities are converted to the appropriate units as required by the question.

Rate in miles per hour = distance in miles / time in hour

to convert 12 minutes to hours; recall that 60 minutes make 1 hour

12 minutes to hour = 12/60 = 0.2hr

to convert 4224 feet to miles; recall 5280 feet is equivalent to 1 mile

4224 feet to miles = 4224/5280 = 0.8 miles

∴ rate = 0.8 / 0.2

rate = 4 miles per hour

the constant rate in miles per hour = 4 miles/hour

4 0
2 years ago
A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50
asambeis [7]

Answer:

The capacitance per unit length is 5.06\times10^{-11}\ F/m

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}

Put the value into the formula

\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}

\lambda=6.068\times10^{-9}\ C/m

We know that,

\lambda=\dfrac{Q}{L}

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

C=\dfrac{\dfrac{Q}{L}}{\Delta V}

C=\dfrac{6.068\times10^{-9}}{120}

C=5.06\times10^{-11}\ F/m

Hence, The capacitance per unit length is 5.06\times10^{-11}\ F/m

7 0
3 years ago
An aluminum cup of 150 cm3 capacity is completely filled with glycerin at 23°C. How much glycerin will spill out of the cup if t
AVprozaik [17]

Answer:

1.19cm^3 of glycerine

Explanation:

Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:

Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum

Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature

coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1

Change in temperature = 41-23 = 18oC

Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3

3 0
2 years ago
A student notices that an inflated balloon gets larger when it is warmed by a lamp. Which best describes the mass of the balloon
nirvana33 [79]
It gets larger because
well let me give you an example
so today in class we looked at a lava lamp with wax inside and there was a lightbulb at the bottom.
we watched as the wax floated up because the molecules inside the wax spreads out and makes the wax less dense.
the wax floats up because (which is related to the balloon getting bigger) the wax is getting less dense and the particles get bigger which ALSO makes the wax less dense.
hope this helps and hope you can relate it to your problem! say thanks if I did help AT ALL! :)
7 0
3 years ago
Read 2 more answers
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