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3 years ago

A 3 N force pushing to the right on a box causes it to accelerate 6 m/s² to the right. What is the mass of the box?

Physics

1 answer:

Aloiza [94]3 years ago

8 0

18 kg would be the answer to that

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A train is moving with a velocity of 70 km/h. If north is taken as the positive direction, what is the direction of movement of

Anna11 [10]

Answer:

+70 km/h north

Explanation:

4 0

2 years ago

Mathphys Help help help

Keith_Richards [23]

Answer:

1030 mph

Explanation:

The new velocity equals the initial velocity plus the wind velocity.

First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

vₓ = 1007 mph

And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1007 mph)² + (231 mph)²

v ≈ 1030 mph

6 0

2 years ago

A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u

skelet666 [1.2K]

Well, if the skydiver is at constant velocity, than there’s no acceleration, as stated by Newton’s first law. Thus the total net force would equate to 0. In order to make this statement true, the answer would have to be exactly 600 N.

7 0

3 years ago

Read 2 more answers

I really need help! Please no one has helped me! Will give BRAINLIEST, THANKS, and RATE!! Please help if you are good at science

adell [148]

I took this test and the correct answer is c
I hope this helps

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3 years ago

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Las condiciones iniciales de un gas son 3000 cm3

slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

$PV=nRT$

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

$n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles$

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

$T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C$

hence, T'=92.70°C

8 0

3 years ago