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3 years ago

Write a haiku

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer:

Explanation:

graphing is helpful

helps visualize the line

of your equation

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Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30

dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0, A + C - F - D = 0

A = F + D - C

Ay = 200 k-N

4 0

3 years ago

Determine the projection (magnitude and sign), or component, of vector v1 along the direction of vector v2. Your answer could be

professor190 [17]

Answer:

- 1.07 ft

Explanation:

V1 = (-5, 7, 2)

V2 = (3, 1, 2)

Projection of v1 along v2, we use the following formula

=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}

So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= $\sqrt{3^{2}+1^{2}+2^{2}}=3.74$

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

Thus, the projection of V1 along V2 is - 1.07 ft.

so we need to find the direction of v2

7 0

4 years ago

A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r

sergij07 [2.7K]

Answer:

The workdone is $W = 1.712 *10^{-20 } \ J$

Explanation:

From the question we are told that

The potential difference is $V = 107 mV = 107 *10^{-3} \ V$

Generally the charge on $Na^{+}$ is $Q_{Na^{+}} = 1.60 *10^{-19 } \ C$

Generally the workdone is mathematically represented as

$W = Q_{Na^{+}}V$

=> $W = 1.60 *10^{-19 } * 107 *10^{-3}$

=> $W = 1.712 *10^{-20 } \ J$

8 0

3 years ago

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma

Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :

$v^2-u^2=2ad$

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{0-(22.22)^2}{2\times 80000}$

$a=-0.00308\ m/s^2$

Value of g, $g=9.8\ m/s^2$

$a=\dfrac{-0.00308}{9.8}\ m/s^2$

$a=(-0.000314)\ g\ m/s^2$

Hence, this is required solution.

8 0

3 years ago

What is the speed of a wave if the wavelength is 100m and the period is 20s

Sergio039 [100]

I don’t think I’m right but I want to say 500 m/s

4 0

4 years ago

Read 2 more answers