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Norma-Jean [14]
2 years ago
6

1. Suppose that you have two samples with the

Physics
1 answer:
grandymaker [24]2 years ago
4 0

the one with has a greater mass will have a higher density because the more particles/mass in a particular space, the more dense it is.

hope i helped you!

mark brainliest please! :D

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Identify two ways in which friction is not useful in science​
kiruha [24]

Answer:

  • Friction between brakes and the wheels of a car or bicycle helps the vehicles slow down.
  • If you do not lubricate your bike regularly with oil, the friction in the chain and axles increases.
8 0
3 years ago
Please can someone give a clear explanantion, <br><br> no extra links thanks
Tems11 [23]

Answer:

the extension recorded by the student would be smaller than the actual extension of the spring

3 0
2 years ago
A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is
Rudik [331]

Answer:

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 × 10²  N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg(h+x)  = 0.5 × k × x²       ...................1

here x is compression in spring

so put here value in equation 1 we get

1.8 × 9.8 × (2.25+x)  = 0.5 × 2× 10² × x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m

3 0
3 years ago
Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu
Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

4 0
2 years ago
The angular velocity (in rpm) of the blade of a blender is given in. If θ=0rad at t=0s, what is the blade's angular position at
Liula [17]

Answer:

θ = 20.9 rad

Explanation:

In a blender after a short period of acceleration the blade is kept at a constant angular velocity, for which we can use the relationship

           w = θ / t

           θ = w t

if we know the value of the angular velocity we can find the angular position, we must remember that all the angles must be in radians

suppose that the angular velocity is w = 10 rpm, let us reduce to the SI system

          w = 10 rpm (\frac{2\pi  rad}{1 rev}) ( \frac{1 min}{60 s} )

= 1,047 rads

let's calculate

          θ = 1,047 20

          θ = 20.9 rad

7 0
3 years ago
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