All categories

3 years ago

Could you help me with this question?

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

Magnitude of static friction force is 70 sin40° = 44.99 N.

No, it is not necessary that it is maximum static friction.

Normal force is equal to 70 cos40° = 53.62 N.

Explanation:

We apply newton law of moton equation along the plane and perpendicular to plane;

Along the plane,

70 sin 40° = $f_{r}$ ---------------(1)

70 cos 40° = N --------------(2)

$f_{r}_{max}$ = μN -----------------(3)

So, it depends on the value of μ that the friction is maximum or not .

You might be interested in

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

3 years ago

The fast server in women's tennis is Venus Williams, who recorded a serve of 130 (209 ) in 2007.

Gemiola [76]

Converting 209 km/hr to m/s:
209 km/hr = 58.0 m/s
Contact distance = 0.1 m

Using the equation of motion:
2as = v² - u², where initial velocity is 0

a = (58)² / (2 x 0.1)
a = 1.68 x 10⁴ m/s²

7 0

4 years ago

A ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit

vichka [17]

As the ball is moving in air as well as we have to neglect the friction force on it

So we can say that ball is having only one force on it that is gravitational force

So the force on the ball must have to be represented by gravitational force and that must be vertically downwards

So the correct FBD will contain only one force and that force must be vertically downwards

So here correct answer must be

<em>Diagram A shows a box with a downward arrow. </em>

8 0

3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

4 years ago

El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La

allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$

En donde:

$\omega_{f}$ : es la velocidad angular final = 2000 rpm = 209,4 rad/s

$\omega_{0}$ : es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.

Resolviendo la ecuación (1) para α, tenemos:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}$

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.

b) El tiempo transcurrido se puede encontrar como sigue:

$\omega_{f} = \omega_{0} + \alpha t$

Resolviendo para t, tenemos:

$t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s$

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

3 0

3 years ago