What is the STANDARD unit of measurement for velocity in Physics?

A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf

makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A

Resposta: 6 m/s^2 e 12N

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6 0

2 years ago

Which solid state component allows a mobile phone to store data even if the battery is removed

Talja [164]

Answer:

Capacitor

Explanation:

Capacitor ; A capacitor is a passive two-terminal electrical component that can store energy in an electric field electrostatically. It works as a small rechargeable battery that stores electricity. However, unlike a battery, it can charge and discharge in the split of a second. Capacitors are widely used to build different types of electronic circuits.

6 0

2 years ago

How to find acceleration without final velocity?

Andrews [41]

Acceleration is the rate of change of a the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. However, if we are not given the final velocity, it would not be possible to use the kinematic equations. One possible to calculate this value would be by generating an equation of distance with respect to time and getting the second derivative of the equation.

7 0

2 years ago

A car drives around a curve with radius 400 m at a speed of 32 m/s. The road is banked at 7.0 degree. The mass of the car is 150

Ronch [10]

Answer:

The magnitude of the centripetal force to make the turn is 3,840 N.

Explanation:

Given;

radius of the cured road, r = 400 m

speed of the car, v = 32 m/s

mass of the car, m = 1500 kg

The magnitude of the centripetal force to make the turn is given as;

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

where;

Fc is the centripetal force

$F_c = \frac{mv^2}{r} \\\\F_c = \frac{(1500)(32)^2}{400}\\\\F_c = 3,840 \ N$

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.

3 0

2 years ago

A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between

Monica [59]

Answer:

a) $t=5.5\mu s$

b) $\Delta t'=12.5\mu s$

Explanation:

a) Let's use the constant velocity equation:

$v=\frac{\Delta x}{\Delta t}$

v is the speed of the muon. 0.9*c
c is the speed of light 3*10⁸ m/s

$t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}$

$t=5.5\mu s$

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

$\Delta t'=\gamma\Delta t$