What is the density of an object that has a mass of 30 g and a volume of 20cm cubed/ to the third power?

leva [86]

Answer:

i would say that the answer would be B

5 0

3 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

3 years ago

How long does it take the lava bomb to reach its maximum height? Answer with three significant digits and the correct unit. A sm

wlad13 [49]

Answer:

The time taken to reach the maximum height is 3.20 seconds

Explanation:

The given parameters are;

The initial height from which the volcano erupts the lava bomb = 64.4 m

The initial upward velocity of the lava bomb = 31.4 m/s

The acceleration due to gravity, g = 9.8 m/s²

The time it takes the lava bomb to reach its maximum height, t, is given by the following kinematic equation as follows;

v = u - g·t

Where;

v = The final velocity = 0 m/s at maximum height

u = The initial velocity = 31.4 m/s

g = The acceleration due to gravity = 9.8 m/s²

t = The time taken to reach the maximum height

Substituting the values gives;

0 = 31.4 - 9.8 × t

∴ 31.4 = 9.8 × t

t = 31.4/9.8 ≈ 3.204

The time taken to reach the maximum height rounded to three significant figures = t ≈ 3.20 seconds

4 0

3 years ago

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle

Artist 52 [7]

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

7 0

3 years ago

The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du

Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

$F_{net} = \sqrt{F_1^2 + F_2^2}$

here given that

$F_1 = 130.0 N$

$F_2 = 4500.0 N$

now we will plug in all data in the above equation

$F_{net} = \sqrt{4500^2 + 130^}$

$F_{net} = 4501.9 N$

so it will have net force 4501.9 N which will be reported by sensor

4 0

3 years ago

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