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monitta
3 years ago
10

A 1040 kg car and 3360 kg truck undergo a perfectly inelastic collision. before the collision, the car was traveling southward a

t 1.80 m/s and the truck westward at 8.25 m/s. m/s. find the velocity (speed and direction) of the wreckage immediately after the collision.
Physics
1 answer:
alexandr1967 [171]3 years ago
6 0
Fddgfjdgkdghsreawoejfkurbgkb,xv biusghorsnj lsjrfgWIL4RGBOHGVkwdc
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Anyone help me please?
choli [55]

Answer:

I  believe that its B my apologies  if its wrong/

Explanation:

3 0
3 years ago
How many types of electrical charge are there in all materials and what are the charges
JulsSmile [24]

There are two types of electric charges; positive and negative

- If you need more info than this let me know

- hope this helps

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
An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz
Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s

Deflection by Earth's Gravity

\Delta =\frac {gt^2}{2}

Now, Time = Distance/Velocity

\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0
3 years ago
A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was
Nookie1986 [14]

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

6 0
3 years ago
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