Ask question

Ask question

All categories

2 years ago

6

An ambulance traveled on roads from the hospital 14 kilometers east, then 16 kilometers north to reach an accident. If the ambul

ance traveled for 0.40 hours, what was its average speed?

1 answer:

Aleksandr-060686 [28]2 years ago

6 0

C . Because I took the test

You might be interested in

If i try to fail and succeed which one did I do

Novay_Z [31]

You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.

7 0

3 years ago

Which units are used to express sound intensity?

grigory [225]

Answer: decibels

Explanation:

4 0

3 years ago

El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión ( $P$ ), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

$P = \frac{F}{A}$

Donde:

$F$ - Fuerza motriz, medida en newtons.

$A$ - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón ( $A$ ), medido en metros cuadrados, es determinado por:

$A=\frac{\pi}{4}\cdot D^{2}$

Donde $D$ es el diámetro del pistón, medido en metros.

Si $D = 0.32\,m$ y $F =17,640\,N$ , entonces la presión neumática es:

$A = \frac{\pi}{4}\cdot (0.32\,m)^{2}$

$A \approx 0.080\,m^{2}$

$P = \frac{17,640\,N}{0.080\,m^{2}}$

$P = 220,500\,Pa$

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

8 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

Spherical mirror is a

madam [21]

Answer:

Spherical mirror is a mirror with a curved reflecting surface.

5 0

2 years ago