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3 years ago

11

if an object is thrown straight up with an initial velocity of 8m/s and takes 3 seconds to strike the ground, from what height w

as the object thrown?

1 answer:

KatRina [158]3 years ago

7 0

Answer:

s = it+1/2 at²

s= 8×3+1/2 (10)(3)²

s = 24+45

s= 69

the object was thrown from a height of 69 meters

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Another unfortunate bug splatters on the windshield of a moving car. Which has the greater change in momentum---the bug or the c

bija089 [108]

Answer:

The change in momentum for the bug and the car will be equal, impulses will be equal in opposite directions and the bug will have a greater acceleration compared to the car, because it has a smaller mass.

Explanation:

Hope this helps..

4 0

3 years ago

"if the left-hand mass is 2.3 kg ,what should the right-hand mass be so that it accelerates downslope at 0.64 m/s2?"

VARVARA [1.3K]

m₁ = 2.3 kg <span>
θ₁ = 70° </span><span>
θ₂ = 17° </span><span>
g = 9.8 m/s²

->The component of the gravitational force on m₁ that is parallel down the incline is: </span><span>
F₁ = m₁ × g × sin(θ₁) </span><span>
F₁ = (2.3 kg) × (9.8 m/s²) × sin(70°) = 21.18 N </span><span>

->The component of the gravitational force on m₂ that is parallel down the incline is: </span><span>
F₂ = m₂ × g × sin(θ₂) </span><span>
F₂ = m₂ × (9.8 m/s²) × sin(70°) = m₂ × (2.86 m/s²) </span><span>

Then the total mass of the system is:
m = m₁ + m₂ </span><span>
m = (2.3 kg) + m₂ </span><span>

If it is given that m₂ slides down the incline, then F₂ must be bigger than F₁, </span><span>
and so the net force on the system must be:
F = m₂×(2.86 m/s²) - (21.18 N) </span><span>

Using Newton's second law, we know that
F = m × a
So if we want the acceleration to be 0.64 m/s², then
m₂×(2.86 m/s²) - (21.18 N) = [(2.3 kg) + m₂] × (0.64 m/s²) </span><span>
m₂×(2.86 m/s²) - (21.18 N) = (1.47 N) + m₂×(0.64 m/s²) </span><span>
m₂×(2.22 m/s²) = (22.65 N) </span><span>
m₂<span> = 10.2 kg</span></span>

5 0

4 years ago

The x-coordinates of two objects moving along the x-axis are given as a function of time (t). x1= (4m/s)t x2= -(161m) + (48m/s)t

Kitty [74]

The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s

The distance between the two objects is
x = x₁ - x₂
= 4t + 161 - 48t + 4t²
x = 4t² - 44t + 161

Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
= 4[ (t - 5.5)² - 5.5² ] + 161
x = 4(t-5)² + 40

Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.

Answer: The distance of the closest approach is 40 m.

5 0

3 years ago

Arrange the following types of photons of electromagnetic radiation in order of decreasing energy: red light, radio, x-rays, γ-r

MrRa [10]

Answer:The correct order is

γ-rays > x-rays> red light > infrared > radio

Explanation:

In the electromagnetic spectrum we have the following electromagnetic radiation Radio waves, infrared radiation, visible light, ultraviolet light, X-ray and gamma rays. Gamma rays have the highest frequency and radio waves the lowest. From the formula;

E=hf, where;

E is energy

h is Planck constant

f is frequency

We can see that the energy of these radiation depends on the magnitude of their frequency. Hence Gamma ray with highest frequency will have the highest energy and radio waves the lowest energy. Red light however is found at the lower end of the visible light spectrum . the correct order is;

γ-rays > x-rays> red light > infrared > radio

8 0

3 years ago

a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

5 0

3 years ago