Can anyone help? Due tomorrow!!

Give an example of something that is cyclic and non - cyclic

LiRa [457]

a clock .. and i guess a non functioning clock ?

7 0

3 years ago

Find the following answer based on the image.

saul85 [17]

We are given:

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

Finding g:

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² (approx)

5 0

3 years ago

Ben hit a 0.25kg nail into a wood with a 5.2kg hammer. If the hammer moves witht the speed of 52 m/s and two fifth of its kineti

Art [367]

Answer:

Explanation:

Mass of nails is 0.25kg

Mass of hammer 5.2kg

Speed of hammer is =52m/s

Then, Ben kinetic energy is given as

K.E= ½mv²

K.E= ½×5.2×52²

K.E= 7030.4J

Given that, two-fifth of kinetic energy is converted to internal energy

Internal energy (I.E) = 2/5 × K.E

Internal energy (I.E) = 2/5 × 7030.4

I.E=2812.16J.

Energy increase is total Kinetic energy - the internal energy

∆Et= K.E-I.E

∆Et= 7030.4 - 2812.16

∆Et= 4218.24J

7 0

3 years ago

vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di

pantera1 [17]

Addition of vectors:

vector u

has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

number.

Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

6 0

3 years ago

A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar

Artist 52 [7]

Answer:

a) v₀ = 32.64 m / s , b) x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

We look for the components of speed with trigonometry

sin 43 = v_{oy} / v₀

cos 43 = v₀ₓ / v₀

v_{oy} = v₀ sin 43

v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

t = x / v₀ cos 43

y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

y = x tan 43 - ½ g x² / v₀² cos² 43

1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

v₀ = √ (35280 / 33.11)

v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

y = $v_{oy}$ t - ½ g t²

.y = v₀ sin43 t - ½ g t²

25 = 32.64 sin 43 t - ½ 9.8 t²

4.9 t² - 22.26 t + 25 = 0

t² - 4.54 t + 5.10 = 0

We solve the second degree equation

t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

t = (4.54 ± 0.46) / 2

t₁ = 2.50 s

t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

x = v₀ₓ t

x = v₀ cos 43 t

x = 32.64 cos 43 2.50

x = 59.68 m

8 0

3 years ago