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3 years ago

The diameter of a marble is 5/8 of an inch. So my question is what is 800 times bigger than a marble?

2 answers:

n200080 [17]3 years ago

5 0

What does "bigger" mean ? ? 800 times the volume of the marble is a diameter of 5.8 inches ... like maybe a softball. 800 times the surface area of the marble is a diameter of 17.7 inches ... like an overinflated basketball. 800 times the radius of the marble is a diameter of 41-ft 8-in ... like a good-sized hot-air balloon.

tangare [24]3 years ago

3 0

Anything that is 500in is 800 times the size of a marble

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the presence of which magnetic feature best explains why a magnet can act a distance on other magnets or on objects containing c

katen-ka-za [31]

Magnetic fields

Explanation:

The presence of magnetic fields best explains why a magnet can act a distance on other magnets or on objects containing certain metals.

Magnetic fields are lines of forces around a bar magnet.
These lines of forces attracts and repels other magnetic bodies and metallic bodies round it.
Magnetic lines of forces originates at the north pole and enters in the south pole.
Areas around a magnetic body are bounded by force fields.
A magnet has permanent magnetic fields round it.

learn more:

Electromagnet brainly.com/question/2191993

#learnwithBrainly

8 0

3 years ago

Pls place the words in the right place I'll give brainless and the points!

Sphinxa [80]

Explanation:

you measure temperature in degrees celsius using a thermometer. Thermal energy is measured in joules. A larger volume of water will take longer to heat up but will store more energy than the smaller object. However, a smaller object will lose it's heat faster than a larger object. A cup of tea has less thermal energy than a swimming pool.

5 0

3 years ago

If the numbers on the plate are 6.0 cm apart, and the spy satellite is at an altitude of 160 km , what must be the diameter of t

Alexxx [7]

Answer:

The diameter of the camera aperture must be greater than or equal to 1.49m

Explanation:

Let the distance separating two objects, x = 6.0 cm = 0.06 m

The distance between the observer and the two objects, d = 160 km = 160000 m

Let ∅ = minimum angular separation between the two objects that the satellite can resolve

tan( ∅) = x/d

Since there is minimum angular separation, tan( ∅) ≈∅

∅ = x/d

∅ = 0.06/160000

∅ = 3.75 * 10⁻⁷rad

For the satellite to be able to resolve the objects,

D ≥ 1.22λ/∅

λ = 560 nm = 560 * 10⁻⁹

D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)

D ≥ 149.33 * 10⁻² m

D ≥ 1.49 m

7 0

3 years ago

A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

PIT_PIT [208]

Their velocity afterward is v=3.467 m/s

Explanation:

Given:

Mass of the first football player= 91.5 kg

Initial velocity of the football player 3.73 m/s

Mass of second football player=63.56 kg

Initial velocity of the second football player=3.09 m/s

To find:

Final velocity of both players=?

Solution:

According to the law of conservation of momentum,

Initial momentum =final momentum

mathematically represented as

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...........................(1)

where

$u_1$ =intial velocity of the football player

$u_2$ = inital velocity of second football player

$v_1$ =finall velocity of the first football player

$v_2$ =final velocity of second football player

after tackling , both the football players moves with the same velocity,

so $v_1=v_2=v$

Hence equation (1) becomes

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

now substituting the values,

$v=\frac{(91.5\times+3.73)+(63.5\times3.09)}{(91.5+63.5)}$ $v=\frac{(341.29+196.210)}{155}$

$v=\frac{537.5}{155}$

v=3.467 m/s

7 0

4 years ago

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

5 0

3 years ago