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ddd [48]
3 years ago
5

What did Dr. Kidwell's fighting so the birds eating a box on different tree trunks

Physics
1 answer:
Gelneren [198K]3 years ago
3 0
I think it's D(Leaves).
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a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin
Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0
4 years ago
Gold-leaf electroscope uses
Kruka [31]
-identifies an electric charge 
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)
7 0
4 years ago
Helps please people
serg [7]
1. The iris regulates the amount of light entering the eye
2. The retina receives and organises visual information
3. The lens refracts light rays in a camera
7 0
2 years ago
Which quantity below is a derived quantity?
Reil [10]

By definition, the speed of an object is given by:

v = \frac{dr}{dt}

Where,

dr/dt: derived from the position with respect to time

Therefore, speed has units of length over units of time.

Thus, speed is a derived quantity, since it depends on the value of two other quantities.

Answer:

a derived quantity is:

C. Speed

4 0
3 years ago
Read 2 more answers
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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