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gizmo_the_mogwai [7]
2 years ago
9

a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?

Physics
1 answer:
IRINA_888 [86]2 years ago
8 0
Vi=15m/s
vf=20m/s
d=50m

a=(vf²-vi²)/2d

average acceleration= 1.75m/s
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Sound waves cannot carry energy through. A water B air C a mirror D a vacuum
Ber [7]
I looked up the question and got D- a vacuum
3 0
2 years ago
A very long wire generates a magnetic field of 0.0020x 10^-4 T at a distance of 10 mm. What is the magnitude of the current? A)
zimovet [89]

Answer:

2*10^-<em>5</em>

Explanation:

<em>B=</em><em>I</em><em>L</em>

<em>I=</em><em>B</em><em>/</em><em>L</em>

<em>I=</em><em>0</em><em>.</em><em>0</em><em>0</em><em>2</em><em>0</em><em>*</em><em>1</em><em>0</em><em>^</em><em>-</em><em>4</em><em>/</em><em>1</em><em>0</em>

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6 0
3 years ago
A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of
Alexandra [31]

Answer:

The maximum no. of electrons- 2.25\times 10^{22}

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

I = \frac{Q}{t}

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

Q = 1.0\times 3600 = 3600\ C

Maximum number of electrons, n is given by:

n = \frac{Q}{e}

where

e = charge on an electron = 1.6\times 10^{- 19}\ C

Thus

n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}

3 0
3 years ago
The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horiz
kotegsom [21]

Answer:

\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

Explanation:

The x- and y- components of the velocity vector can be written as following:

\vec{v}_x = ||\vec{v}||\cos(\theta)\^i

\vec{v}_y = ||\vec{v}||\sin(\theta)\^j

Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

3 0
2 years ago
A solid box made from a material whose density is rhoS floats with two thirds of its volume submerged in a liquid whose density
jok3333 [9.3K]

Answer:

\frac{\rho_S}{\rho_L} = \frac{2}{3}

Explanation:

density of the solid box material = \rho_s

density of the liquid material = \rho_L

Given that

solid box floats with two thirds of its volume submerged in a liquid

let V be the volume of the box

then,

V\rho_sg= \frac{2V}{3}\rho_L g

⇒\frac{\rho_S}{\rho_L} = \frac{2}{3}

so, the ratio of densities of solid and and the liquid is 2/3

7 0
3 years ago
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