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Nataly_w [17]
3 years ago
7

A _____ lens is used to converge light A. Convex B. Concave

Physics
2 answers:
Vanyuwa [196]3 years ago
7 0

Answer:

convex

Explanation:

disa [49]3 years ago
6 0
A convex lens<span> makes </span>light<span> rays </span><span>converge </span>
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Which chemical equation correctly shows the formation of water from hydrogen
nekit [7.7K]

Answer:

d. H + O2 ------ H2O

that's the formation of water

6 0
3 years ago
Read 2 more answers
What is the weight of a 4.2 kg bowling ball on Mars?
Nataliya [291]

What is the weight of a 4.2 kg bowling ball on Mars?

Answer:

1.59 kg

Explanation:

The formula is:

<u>F = G((Mm)/r2) </u>

F is the gravitational force between two objects,

G is the Gravitational Constant (6.674×10-11 Newtons x meters2 / kilograms2),

M is the planet's mass (kg),

m is your mass (kg), and

r is the distance (m) between the centers of the two masses (the planet's radius).

Hope this helps

--Jay

8 0
4 years ago
(a) If a long rope is hung from a ceiling and waves are sent. up the rope from its lower end, why does the speed of the waves ch
vazorg [7]

My response to question (a) and (b) is that all of the element of the rope need to  aid or support the weight of the rope and as such, the tension will tend to increase along with height.

Note that It increases linearly, if the rope is one that do not stretch. So, the wave speed v= √ T/μ   increases with height.

<h3>How does tension affect the speed of a wave in a rope?</h3>

The Increase of the tension placed on a string is one that tends to increases the speed of a wave, which in turn also increases the frequency of any given length.

Therefore,  My response to question (a) and (b) is that all of the element of the rope need to  aid or support the weight of the rope and as such, the tension will tend to increase along with height.  Note that It increases linearly, if the rope is one that do not stretch. So, the wave speed v= √ T/μ   increases with height.

Learn more about tension from

brainly.com/question/2008782

#SPJ4

See full question below

(a) If a long rope is hung from a ceiling and waves are sent up the rope from its lower end, why does the speed of the waves change as they ascend? (b) Does the speed of the ascending waves increase or decrease? Explain.

4 0
2 years ago
A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
    = 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
             = 4,500,000 m/s²
But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
6 0
4 years ago
We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth
Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = \frac{m}{m+M} \ v_o

for when n balls have collided

          v = \frac{m}{n \ m + M}  v₀

Now let's analyze the case of the bouncing ball (elastic)

     

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = \frac{m-M}{m + M}   v₀

we substitute in equation 1 to find v₂

            \frac{M}{m}  v₂ = v₀ -  \frac{m-M}{m+M}   v₀

            v₂ = \frac{m}{M}  ( 1 - \frac{m-M}{m+M} ) \ v_o

            v₂ = \frac{m}{M}  ( \frac{2M}{m+M} ) \ \ v_o

            v₂ = \frac{2m}{m +M}  \ v_o  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0
3 years ago
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