An object is 2.0 m in front of a plane mirror. Its image is:?

A 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal exp

Gnom [1K]

Answer:

a.) 1567.2 m/s

b.) 149.4 m/s

Explanation:

Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.

The x-component of the third part can be calculated by assuming that it moves in a positive x axis.

The third mass = 26 - ( 7.8 + 8.8)

The third mass = 26 - 16.6

The third mass = 9.4kg

since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion

26 x 350 = -8.8 x 640 + 9.4V

9100 = -5632 + 9.4V

9.4V = 9100 + 5632

9.4V = 14732

V = 14732/9.4

V = 1567.2 m/s

(b) y-component of the velocity of the third part will be

7.8 x 180 = 9.4 V

1404 = 9.4V

V = 1404/9.4

V = 149.4 m/s

7 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago

Object a travels in the +x-direction before hitting a stationary object

Leto [7]

The object’s resultant angle of motion with the +x-axis after the collision is 47°

<span>From object A:

1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.

Now, we know, tan</span>Ф = $\frac{y}{x}$

⇒tanФ = $\frac{6.2 × 10^4 }{5.7 × 10^4}$

⇒tanФ = 1.088

⇒ Ф = $tan^{-1}$ 1.088
= 47.4 ≈ 47

8 0

3 years ago

Read 2 more answers

In order to change a 2.15 kg handcart's velocity from 1.2 m/s

photoshop1234 [79]

Answer: E 1.29 N•s

Explanation: math

7 0

3 years ago