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3 years ago

An observation is received directly or indirectly by

Physics

2 answers:

Roman55 [17]3 years ago

6 0

Answer:

D. All of the above.

Explanation:

An observation is received directly or indirectly by all of the above. When observing things, you need to visually be able to see what is being observed, and be able to touch hear and even taste. This ensures that you can get the best out of the observation.

sveta [45]3 years ago

4 0

Answer:

All of the above

Explanation:

because these are all senses of the body and therefore you're receiving signals from all of them all the time

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6. A 7-kg bowling ball moving at 6.0 m/s strikes a 1-kg bowling pin. 11 we van

brilliants [131]

Answer:

210 N

Explanation:

The change in momentum of the ball is equal (in magnitude) to the impulse given to the pin, according to

$\Delta p = F \Delta t$ (1)

where

$\Delta p$ is the change in momentum of the ball

F is the force exerted on the pin

$\Delta t$ is the duration of the collision

For the bowling ball, we have

m = 7 kg (mass)

u = 6.0 m/s (initial velocity)

v = 4.5 m/s (final velocity)

So, the change in momentum (in magnitude) is

$\Delta p = m(u-v)=(7)(6.0-4.5)=10.5 kg m/s$

We also know the duration of the collision,

$\Delta t = 0.05 s$

And so if we re-arrange eq.(1), we find the force exerted on the pin:

$F=\frac{\Delta p}{\Delta t}=\frac{10.5}{0.05}=210 N$

3 0

3 years ago

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy

masha68 [24]

Answer:

A. Both the girl and the boy have the same nonzero angular displacement.

Explanation:

Since girl and boy both are standing on the same merry go round so here we can say that boy and girl both will revolve with same angular speed as the speed of the merry go round

so the angular displacement is given as

$\theta = \omega t$

so here since both has same angular speed that is the speed of the disc so they both will turn by same angle in the same interval of time

So we have correct answer as

A. Both the girl and the boy have the same nonzero angular displacement.

4 0

4 years ago

The parent function `f\left(x\right)=\sqrt[3]{x}` is compressed vertically by a factor of `\frac{1}{3}` and then translated 3 un

tensa zangetsu [6.8K]

Answer:

The answer is below

Explanation:

Given that f(x) = x√3.

A function can be vertically stretched or compressed by multiplying it by a positive constant. If the constant is greater than 1, it is vertically stretched and if the constant is less than 1 it is vertically compressed.

If a function f(x) = x is compressed or stretched by a constant a, then the new function g(x) = a f(x)

If a function f(x) = x is translated a units down, then the new function g(x) = f(x) - a

If a function f(x) = x is translated a units left, then the new function g(x) = f(x-a)

If f(x) = x√3 is compressed vertically by a factor of 1/3. The new function is

$f(x)'=x\sqrt{3} *\frac{1}{3} \\\\f(x)'=\frac{x}{3} \sqrt{3}$

If it is then translated 3 units left and 7 units down, the transformed function g(x) is:

$g(x)=(\frac{x-3}{3}\sqrt{3} )-7$

4 0

3 years ago

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-hi

Andrew [12]

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is

k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

3 0

3 years ago

Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The

Mnenie [13.5K]

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

Answer: So, the final angular momentum is 27.73 revolution per minute

7 0

3 years ago