Two examples of inventions that increase friction?

Vector B has x, y, and z components of 2.2, 6.6, and 5.5 units, respectively.

Aloiza [94]

Answer:

8.9 units

Explanation:

The magnitude of a 3-D vector B can be calculated by using the formula:

$|B| = \sqrt{B_x^2+B_y^2+B_z^2}$

where $B_x, B_y, B_z$ are the x, y and z components of the vector, respectively.

For the vector in the problem:

$B_x = 2.2\\B_y = 6.6\\B_z = 5.5$

Substituting into the equation, we find the magnitude of B:

$B=\sqrt{2.2^2+6.6^2+5.5^2}=8.9$

So, the magnitude of B is 8.9 units.

3 0

3 years ago

The speed of sound is 344 m/sec when the air is 20 degrees Celsius how far is the source of the sound if it takes 8 seconds for

pishuonlain [190]

Answer: 2752 m

Explanation:

344*8=

5 0

2 years ago

The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo

mylen [45]

You can just use basic trigonometry to solve for the x & y components.

vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

6 0

2 years ago

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

butalik [34]

Answer:

a) I = -2257.6 Kg*m/s

b) F = -451,520N

Explanation:

part a.

we know that:

I = $P_f-P_i$

where I is the impulse, $P_f$ the final momentum and $P_i$ the initial momentum.

so:

I = $MV_f-MV_i$

where M is the mass, $V_f$ the final velocity and $V_i$ the initial velocity.

Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.

now, we will use the law of the conservation of energy:

$E_i=E_f$

so:

mgh = $\frac{1}{2}MV_i^2$

where g is the gravity and h the altitude. So, replacing values, we get:

(85kg)(9.8m/s^2)(36m)= $\frac{1}{2}(85kg)V_i^2$

solving for $V_i$ :

$V_i = 26.56m/s$

Then, replacing in the initial equation:

I = $MV_f-MV_i$

I = $(85kg)(0m/s)-(85kg)(26.56m/s)$

I = -2257.6 Kg*m/s

Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.

part b.

we know that:

Ft = I

where F is the average force, t is the time and I is the impulse. So, replacing values, we get:

F(0,005s) = -2257.6 Kg*m/s

solving for F:

F = -451520N

Finally, the force is -451,520N, it is negative because it is upwards.

3 0

3 years ago

Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

snow_lady [41]

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength $\lambda=0.06m$ .

The frequency $f$ of the waves is 14.8 waves every 3 seconds or

$f=14.8/3 =4.33Hz$ .

Now the relationship between wavelength $\lambda$ , frequency $f$ and speed $v$ of the waves is:

$v=\lambda f$

We put in the values $\lambda=0.06m$ and $f=4.933Hz$ and get:

$\boxed{v=0.06*4.922=0.296m/s}$

Now the period $T$ is just the inverse of the frequency, or

$T=\frac{1}{f}$

$\boxed{T=\frac{1}{4.933}=0.203\:seconds }$

4 0

3 years ago