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3 years ago

PLEASE ANSWER QUICKILY

Physics

2 answers:

velikii [3]3 years ago

8 0

Answer is less solar radiation

Kobotan [32]3 years ago

6 0

Answer:

Water retains less solar radiation than land.

Explanation:

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What are the main differences between the manual, oscillate and pulse functions?

gladu [14]

Answer:

Manuel function can be used when we move it. Oscillate function automatically moves on its own. The pulse function mimics the pulse of a heart rate and can only be used by pushing the button.

Explanation:

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3 years ago

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Albert is on Earth and measures the speed of light from the sun as 3 E8 m/s. He then gets in a spaceship and speed toward the su

Flura [38]

Choice B, 3 E8 m/s because it will stay the same no matter where he is or how fast he's going

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3 years ago

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The graph below shows the composition of the air we breathe. Which of the following statements correctly classified the air we b

levacccp [35]

Answer:

D. The air we breathe is a homogeneous mixture because it has substances but the same appearance throughout.

Explanation:

The air we breathe is a homogeneous mixture.

Air is a mixture of many gases. A homogeneous mixture is one whose constituents exists in one phase.

Air is a gas - gas mixtures.

A mixture has indefinite composition. They consist of two or more elements and or compounds in any proportion by mass.

Constituents retain their identities i.e. physical properties are retained.

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3 years ago

A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi

Marrrta [24]

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, $\mu = 0.19$

Now,

We know that the circumference of the circle is given by:

C = $2\pi R$

27 = $2\pi R$

$R = \frac{27}{2\pi } = 4.29\ in$

Also, in case of rolling, we know that:

Angular velocity, $\omega = \frac{v}{R}$

$v = \omega R$

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

$m\omega' = m\omega + I\omega$

Moment of Inertia, I = $\frac{2}{5}mR^{2} = mk^{2}$

$mv'R = mvR + mk^{2}\times \frac{v}{R}$

$v'R = vR + k^{2}\times \frac{v}{R}$

$276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}$

$1184.04 - 756.903 = 4.29v$

v = 99.56 in/s

Now,

Friction force, f = $\mu mg$

Also, acceleration of the ball can be computed as:

$\mu mg = - ma$

$a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}$

Now, the distance, d covered by the ball before rolling without slipping:

$v^{2} = v'^{2} + 2ad$

$99.56^{2} = 276^{2} + 2\times (- 73.357)d$

$99.56^{2} - 276^{2} = 2\times (- 73.357)d$

d = 451.65 in = 37.53 ft

7 0

3 years ago

A 3,162-kg truck moving with a velocity of 12 m/s to the East hits a 510-kg parked car. The impact causes the car to be set in m

Musya8 [376]

Given:

The mass of the truck is m1 = 3162 kg

The speed of the truck is v1i = 12 m/s in East

The mass of the parked car is m2 = 510 kg

The speed of car is v2i = 0 m/s

The speed of car after collision is v2f = 24 m/s in East

To find the speed of the truck after collision.

Explanation:

The final velocity of the truck will be

$\begin{gathered} m1v1i+m2v2i=m1v1f+m2v2f \\ v1f\text{ =}\frac{m1v1i-m2v2f+m2v2i}{m1} \\ =\frac{3162\times12-510\times24+510\times0}{3162} \\ =8.129\text{ m/s} \end{gathered}$

Thus, the speed of the truck after collision is 8.129 m/s

6 0

1 year ago