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2 years ago

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Complete the statements using data from Table C of your Student Guide. The speed of the 2 kg cart after 5 seconds is cm/s. The s

peed of the 4 kg cart after 3 seconds is cm/s. The speed of the 6 kg cart after 10 seconds is cm/s.

2 answers:

Stels [109]2 years ago

8 0

Answer:

Complete the statements using data from Table C of your Student Guide.

The speed of the 2 kg cart after 5 seconds is

80.0

The speed of the 4 kg cart after 3 seconds is

24.0

The speed of the 6 kg cart after 10 seconds is

53.3

ArbitrLikvidat [17]2 years ago

4 0

Answer:

80.0 cm/s

24.0 cm/s

53.3 cm/s

Explanation:

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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

Blizzard [7]

<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity $V_{esc}$ is given by the following equation:

$V_{esc}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M$ is the mass of the asteroid

$R$ is the radius of the asteroid

On the other hand, we know the density of the asteroid is $\rho=3.84(10)^{8}g/m^{3}$ and its volume is $V=2.17(10)^{12}m^{3}$ .

The density of a body is given by:

$\rho=\frac{M}{V}$ (2)

Finding $M$ :

$M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})$ (3)

$M=8.33(10)^{20}g=8.33(10)^{17}kg$ (4) This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

$V=\frac{4}{3}\piR^{3}$ (5)

Finding $R$ :

$R=\sqrt[3]{\frac{3V}{4\pi}}$ (6)

$R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}$ (7)

$R=8031.38m$ (8) This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

$V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}$ (9)

Finally:

$V_{esc}=117.626m/s$ This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

The process of changing a property of a wave to transmit information is called .

BARSIC [14]

Answer : The process of changing a property of a wave to transmit information is called Modulation.

Explanation :

Modulation is the process of changing the property of wave to transmit information. This is done with the help of modulator.

In modulation, the message signal is superimposed on a high frequency signal. A sine wave ( usually high frequency ) is used as a high frequency carrier wave.

Modulation can be done in many ways like :

(1) Frequency modulation

(2) Amplitude modulation

(3) Pulse modulation

8 0

2 years ago

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A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H

ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

$M=\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o = 2.0 m$ is the height of the real object (the man)

$q=50 mm =0.050 m$ is the distance of the image from the lens

$p = 5.0 m$ is the distance of the object (the man) from the lens

Solving the formula for $h_i$ , we find

$h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm$

And the negative sign means the image is inverted.

6 0

2 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

7 0

2 years ago