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3 years ago

can you help me create a sketch of two different objects with one that has a greater density than the other?

1 answer:

6 0

I don't know how good you are at sketching ... I'm terrible.
But you can put the point across in a dramatic way if you
can sketch a bowling ball and a basketball ... you'll need
to clearly identify them with the markings you sketch on
each ball.

They're the same shape and nearly the same size, but
there's a huge difference in their densities.

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A golfer hits a golf ball at an angle of 25 degrees to the ground at a speed of 76 m/s. If the gold ball covers a horizontal dis

Lady bird [3.3K]

The ball's horizontal position x and vertical position y at time t are given by

x = (76 m/s) cos(25º) t

y = (76 m/s) sin(25º) t - 1/2 g t²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball's initial vertical velocity is (76 m/s) sin(25º) ≈ 32.12 m/s.

Its initial horizontal velocity is (76 m/s) cos(25º) ≈ 68.88 m/s.

The ball stays in the air for as long as y > 0. Solve y = 0 for t :

(76 m/s) sin(25º) t - 1/2 g t² = 0

t ((76 m/s) sin(25º) - 1/2 g t ) = 0

t = 0 or (76 m/s) sin(25º) - 1/2 g t = 0

Ignore the first solution.

(76 m/s) sin(25º) - 1/2 g t = 0

(76 m/s) sin(25º) = (4.90 m/s²) t

t = (76 m/s) sin(25º) / (4.90 m/s²)

t ≈ 6.55 s

Recall that

v² - u² = 2 a ∆y

where u and v denote initial and final velocities, a is acceleration, and ∆y is displacement. At maximum height, the ball has zero vertical velocity, and taking the ball's starting position on the ground to be the origin, ∆y refers to the maximum height. So we have

0² - ((76 m/s) sin(25º))² = 2 (-g) ∆y

∆y = ((76 m/s) sin(25º))² / (2g)

∆y ≈ 52.6 m

5 0

3 years ago

If a wave were to hit a flexible, moveable surface what will it do?

Vika [28.1K]

what ever force the wave hit it at (on the x-axis) would create another wave from the force of the first wave. though the wave would be smaller because of other forces

4 0

4 years ago

Read 2 more answers

In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of

Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, $a_{Net}$ = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - $a_{Net}$ × t

∴ t = (v₁ - v₂)/ $a_{Net}$ = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2· $a_{Net}$ ·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0

3 years ago

When two or more forces act together on an object what do they combine to form

Daniel [21]

Without knowing anything about their magnitudes or directions, the only thing you can always say about thier combination is that it's the "net force" on the object.

7 0

4 years ago

Two objects must be in contact for them to exert a force on each other. True False

Irina-Kira [14]

Answer:

False

Explanation:

The given statement "Two objects must be in contact for them to exert a force on each other" is not true as there are many types of forces that doesn't require being in contact for exerting a force.

One such example is the gravitational force acting between two bodies. Gravitational force is the force of pull with which a body pulls another body without being in contact.

For two bodies of masses 'M' and 'm' separated at a distance of 'R', the gravitational force is given as:

$Force=\frac{GMm}{R^2}\\Where,G\to \textrm{Universal Gravitational constant}$

The gravitational force acts always act between bodies that have mass. The bodies are not in contact yet experience force.

Therefore, the given statement is false.

5 0

4 years ago