Ask question

Ask question

All categories

3 years ago

15

a car going 22 m/s accelerates to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration of the ca

r

1 answer:

Maksim231197 [3]3 years ago

5 0

Using the formula

$a\: = \frac{v - u}{t}$

$a = \frac{35 m/s \: - \: 22 m/s}{5s}$

$a = 2.6 {m/s^2}$

You might be interested in

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

3 years ago

A refrigerator is used to remove 84 kj/min of heat from a tank. If the electric power consumed by the refrigerator is 1. 2 kw, w

zvonat [6]

Answer:

84 kj/min = 1.4 kj/sec

Power Out / Power In = Heat Out / Heat In - Coefficient of Performance

1.4 kj/sec / 1.2 kj/sec = 1.17 = COP

3 0

3 years ago

si un disco fue lanzado con una fuerza de 1000 N, y el disc recorrió una distancia e o.6 m ¿Qué trabajo efectuó el lanzador ?

Alexeev081 [22]

Answer:

Trabajo realizado = 600 Nm

Explanation:

Dados los siguientes datos;

Fuerza = 1000 Newton

Distancia = 0.6 metros

Para encontrar el trabajo realizado;

$Trabajo \; realizado = fuerza * distancia$

Sustituyendo en la ecuación, tenemos;

$Trabajo \; realizado = 1000 * 0.6$

Trabajo realizado = 600 Nm

4 0

2 years ago

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

VLD [36.1K]

Answer:

The the maximum emf is $2.52\times10^{-11}\ V$

Explanation:

Given that,

Magnetic field $B = 1.40\times10^{-3}\ T$

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

$\epsilon=NBA\omega$

Where, N = number of turns

B= magnetic field

A = area

Put the value in to the formula

$\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60$

$\epsilon=2.52\times10^{-11}\ V$

Hence, The the maximum emf is $2.52\times10^{-11}\ V$

5 0

3 years ago

A radioactive isotope has a life for 15 minutes. It has an initial count rate of 36000 Bg. What will the count be after 1 hour?

lukranit [14]

Answer:

18000 Bg

Explanation:

after 15 minutes = 36000/2

After 30 minutes = 36000/4

30 minutes × 2 = 1 hour

After 1 hour = 36000/4 × 2 = 18000 Bg

3 0

3 years ago