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2 years ago

If in 30 minutes Allex runs to a store that is 4 km away, what is her average speed in km/h? km/h

Physics

2 answers:

dlinn [17]2 years ago

6 0

Im pretty sure the answer is 8km/h

grin007 [14]2 years ago

5 0

Average speed is given as the ratio of total distance traveled in a particular time interval.

t = time of travel for allex = 30 min = 30/60 hour = 0.5 hour (Since 1 hour = 60 min)

d = distance traveled in time "t" = 4 km

V = average speed of allex

Average speed is given as

V = d/t

V = 4/0.5

V = 8 km/h

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A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

Vaselesa [24]

Answer:

$1.57 * 10^{3} Q$

Explanation:

The volume charge density is defined by ρ = $\frac{Q}{V}$ (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are $\frac{Coulombs}{m^{3} }$ , so we have to express the radius in meters:

inner radius = $4 cm * \frac{1 m}{100 cm} = 0.04m$

outer radius = $6 cm * \frac{1m}{100cm} = 0.06m$

Now, we know that the volume of the sphere is calculated by the formula:

$V = \frac{4}{3}\pi r^{3}$ , and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = $\frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3})$ , where $r_{o}$ is the outer radius and $r_{i}$ is the inner radius.

Replacing the volume formula in the Equation A:

ρ = $\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}$

ρ = $\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }$

Replacing the values of the outer and inner radius whe have:

ρ = $\frac{3Q}{4\pi (1.52 * 10^{-4})}$

ρ = $1.57 * 10^{3} Q$

4 0

2 years ago

A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has ra

kaheart [24]

Answer:

The plank moves 0.2m from it's original position

Explanation:

we can do this question from the constraints that ,

the wheel and the axle have the same angular speed or velocity
the speed of the plank is equal to the speed of the axle at the topmost point .

thus ,

since the wheel is pure rolling or not slipping,

⇒ $v=wr$

where

$v$ - speed of the wheel

$w$ - angular speed of the wheel

$r$ - radius of the wheel

since the wheel traverses 1 m let's say in time ' $t$ ' ,

$v_{w}=\frac{distance}{time} =\frac{1}{t}$

∴

⇒ $w=\frac{v_{w}}{r} = \frac{1}{t*0.25}$

the speed at the topmost point of the axle is :

⇒ $v_{a}=w*r\\v_{a}=\frac{1}{t*0.25} *0.05\\v_{a}=\frac{1}{5t}$

this is the speed of the plank too.

thus the distance covered by plank in time ' $t$ ' is ,

⇒ $d=v_{a}*t\\d=\frac{1}{5t} *t\\d=\frac{1}{5} = 0.2m$

8 0

2 years ago

A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent

SCORPION-xisa [38]

Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec

Solving for ωf, we get:

ωf = -0.84 rad/sec (clockwise)

5 0

2 years ago

I need the answer a sap please

pogonyaev

D is the answer
I hope it helps!

7 0

2 years ago

Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? W

Akimi4 [234]

Answer:

Same direction to produce maximum magnitude and opposite direction to produce minimum magnitude

Explanation:

Let a be the angle between vectors A and B. Generally when we add A to B, we can split A into 2 sub vectors, 1 parallel to B and the other perpendicular to B.

Also let A and B be the magnitude of vector A and B, respectively.

We have the parallel component after addition be

Acos(a) + B

And the perpendicular component after addition be

Asin(a)

The magnitude of the resulting vector would be

$\sqrt{(Acos(a) + B)^2 + (Asin(a))^2}$

$= \sqrt{A^2cos^2a + B^2 + 2ABcos(a) + A^2sin^2a}$

$= \sqrt{A^2(cos^2a + sin^2a) + B^2 + 2ABcos(a)}$

$= \sqrt{A^2 + B^2 + 2ABcos(a)}$

As A and B are fixed, the equation above is maximum when cos(a) = 1, meaning a = 0 degree and vector A and B are in the same direction, and minimum with cos(a) = -1, meaning a = 180 degree and vector A and B are in opposite direction.

8 0

2 years ago