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LekaFEV [45]
4 years ago
6

A car travelling 10m/s increases its speed to 20m/s over a time of 4s. What is the acceleration of the car? A car is traveling a

t a speed of 15m/s. It experiences an acceleration of -3m/s^2, that lasts for 4 seconds. What is the final velocity of the car? I need an expanation on HOW to solve this.
Physics
1 answer:
Alexxx [7]4 years ago
6 0

Answer:

Explanation:

Acceleration is the change in velocity with respect to time.

Acceleration = change in velocity/Time

Acceleration = final velocity - initial velocity/Time

Given initial velocity = 10m/s

Final velocity = 20m/s

Time taken = 4s

Acceleration = 20-10/4

Acceleration = 10/4

Acceleration =2.5m/s²

For the second part of the question:

Given parameters

initial velocity = 15m/s

acceleration = -3m/s²

time = 4seconds

a = v-u/t

-3 = v-15/4

cross multiply

-12 = v-15

add 15 from both sides

-12+15 = v-15+15

3 = v

<em>Hence the final velocity of the car is 3m/s</em>

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A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
vovangra [49]

Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

donde v0 es la velocidad inicial del auto = 28.7m/s

v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

                 t   = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

6 0
3 years ago
Plz answer this! I am stumped
Gwar [14]
MgCl2
Mg = magnesium
Cl = chlorine

Magnesium + chlorine = magnesium chloride.

This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.

So ur answer would be magnesium chloride. :)
6 0
4 years ago
An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object
EleoNora [17]

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

5 0
3 years ago
Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p
Alexxandr [17]

Answer:

magnitude of A − B =  15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a  magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = \sqrt{(-15)^{2}+(-5)^{2}}

|A - B| = \sqrt{250}

|A - B| = 15.81 km

7 0
3 years ago
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