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2 years ago

Express 75−−√ in the form ab√.

1 answer:

8 0

We can write 75 = 25 x 3
and √75 = √ (25 x 3) = √25 x √3
Now we know that square root of 25 is 5 a 5x5 = 25
so we can express this in the form of 5√3.

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Which law of motion accounts for the following statement? "When a marble and a billiard ball are impacted by the same force, the

Komok [63]

The second law explains this.

4 0

2 years ago

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A stiff wire 50.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l

zloy xaker [14]

Answer:

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

Explanation:

The stiff wire 50.0cm long bent at a right angle in the middle

One section lies along the z axis and the other is along the line y=2x in the xy plane

$\frac{y}{x} = 2$

tan θ = 2

Therefore,

slope m = tan θ = y / x

$\theta=\tan^-^1(2)=63.4^0$

Then length of each section is 25.0cm

so, length vector of the wire is

$\hat I= (-25.0)\hat k +(25.0) \cos 63.4^0 \hati +(25.0) \sin63.4^0 \hatJ\\\\\hat I = (11.2) \hat i + (22.4) \hat j - (23.0) \hat k$

And magnetic field is B = (0.318T)i

Therefore,

$\bar F = \hat I (\bar l \times \bar B)$

$\bar F = (20.0)[(0.112m)i +(0.224m)j-(0.250m)k \times 90.318T)i]$

$= (20.0)(i(0)+j(-0.250)(0.318T)+k[0-(0.224m)(0.318T)]\\\\=(20.0)(-0.250)(0.318)j-(20.0)(0.224)(0.318T)\\\\=-(1.59N)j-(1.425N)k$

Magnitude of the force is

$F = \sqrt{(-1.59N)^2+(-1.425N)^2\\} \\F = 2.135N$

Direction is

$\alpha = \tan^-^1(\frac{-1.425N}{-1.59N} )\\\\= 41.8^0$

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

5 0

2 years ago

Read 2 more answers

Water, of density 1000 kg/m3, is flowing in a drainage channel of rectangular cross-section. The width of the channel is 15 m, t

Rom4ik [11]

Answer:

Water flowing rate= (300000kg/s) = (300000l/s)

Explanation:

First with the section of the channel, the depth of the water and the speed of the fluid we can determine the volume of fluid that circulates per second through the channel:

Volume per time= 15m × 8m × (2.5m/s)= 300 m³/s

With this volume of circulating fluid per second elapsed, we multiply it by the density of the water to determine the kilograms or liters of water that circulate through the channel per second elapsed:

Water flowing rate= (300m³/s) × (1000kg/m³)= (300000kg/s) = (300000l/s)

Taking into account that 1kg of water is approximately equal to 1 liter of water.

8 0

2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

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

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

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

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

6 0

2 years ago

An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

$v_{x}$ = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

$t = \frac{X}{v_{x}}$

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel = 3.67 x 10⁻⁸ sec

$v_{y}$ = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = $v_{y}$ t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0

2 years ago