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3 years ago

Alex collides with a desk as he runs through his office. If the combined

Physics

2 answers:

Ivan3 years ago

8 0

Answer:

400 kg.m/s

Explanation:

In a closed system, the principle of conservation of momentum applies. It means. the total momentum before collision equals to the total momentum after collision hence since the initial momentum is given as 400 kg.m/s, the final sum will be the same.

Alexus [3.1K]3 years ago

8 0

Answer:

400 kg•m/s north

Explanation:

answer for quiz from apex

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A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o

AlekseyPX

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

$P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB$ (1)

so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

$P_{out} = 0.100 W = 100 mW$

3 0

4 years ago

An ideal spring obeys Hooke’s law, F~ = −k~x. A mass of 0.50 kilogram hung vertically from this spring stretches the spring 0.07

masha68 [24]

The spring constant is 66.7 N/m

Explanation:

First of all, we have to find the magnitude of the force acting on the spring. This is equal to the weight of the mass hanging on the spring, which is:

$F=mg$

where:

m = 0.50 kg is the mass of the object

$g=10 m/s^2$ is the acceleration of gravity

Substituting,

$F=(0.50)(10)=5 N$

Now we can use Hookes' law to find the constant of the spring:

$F=-kx$

where

F is the force applied

k is the spring constant

x is the stretching of the spring

Here we have:

F = 5 N

While the stretching is

x = 0.075 m

Therefore, ignoring the negative sign in the formula (which only tells us the direction), we find the spring constant:

$k=\frac{F}{x}=\frac{5}{0.075}=66.7 N/m$

#LearnwithBrainly

5 0

3 years ago

In an experiment all participants receive a treatment but those in the control group receive a false treatment which is called a

Alchen [17]

Answer:

Placebo ?

Explanation:

6 0

3 years ago

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

5 0

3 years ago

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.500 mm. This results in an interference pat

ipn [44]

Answer:

0.000109375 m

Explanation:

d = Distance between grating = 0.5 mm

m = Order

$\lambda_1=\dfrac{d}{8}$

Minima relation

$m-\dfrac{1}{2}$

For fourth order minima

$dsin\theta=(4-\dfrac{1}{2})\lambda_1$

For second maxima

$dsin\theta=2\lambda_2$

From the two equations we get

$(4-\dfrac{1}{2})\lambda_1=2\lambda_2\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\lambda_1}{2}\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\dfrac{0.5\times 10^{-3}}{8}}{2}\\\Rightarrow \lambda_2=0.000109375\ m$

The wavelength is 0.000109375 m

7 0

4 years ago