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4 years ago

The weight of a box is found to be 30 N. What is the approximate mass of the box?

Physics

2 answers:

svlad2 [7]4 years ago

6 0

Answer:

3kg

Explanation:

3kg is the appropriate mass

gtnhenbr [62]4 years ago

5 0

Answer:

3kg

Explanation:

just took the test

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A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find

uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

$E_{k1}=\frac{1}{2}*m*v1^{2}$

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0

3 years ago

What is the gravitational acceleration close to the surface of a planet with a mass of 9ME and radius of 3RE, where ME and RE ar

Papessa [141]

Answer:

9.78 m/s²

Explanation:

To solve this, we use the gravitational formula

g = GM/r², where

g = acceleration due to gravity

G = gravitational constant

M = mass of the planet

r = radius of the planet

From the question, we got that the mass of the planet is

M = 9ME, where ME = 5.95*10^24

M = 9 * 5.95*10^24

M = 5.355*10^25 kg

Also, the Radius of the planet, R = 3RE, where RE = 6.37*10^6

R = 3 * 6.37*10^6

R = 1.911*10^7 m

On applying the values of both R and M to the equation, we get

g = GM/r²

g = (6.67*10^-11 * 5.355*10^25) / (1.911*10^7)²

g = 3.57*10^15/3.65*10^14

g = 9.78 m/s²

Therefore, the acceleration due to gravity on the planet is 9.78 m/s²

Please vote brainliest if it helped you <3

5 0

3 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago

A car is running at a velocity of 50 miles per hour and the driver accelerates the car by 10 miles per hour square.How far the c

krek1111 [17]

It will be 80 miles and it can be done only in 16 min

7 0

3 years ago

Read 2 more answers

Find the mass of a sample of water if its temperature dropped 24.8 °C when it lost 870 J of

rosijanka [135]

Answer:

8.35 × $10^{-3}$ kg

Explanation:

Remember the formula for heat:

$Q=mc\Delta T$

You just need to solve for m, doing so you're left with:

$m=\dfrac{Q}{c\Delta T}$

Replace the variables with numbers and that's it.

7 0

2 years ago