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2 years ago

As an airplane leaves the runway and goes into the air, has acceleration occurred?

1 answer:

6 0

A, yes because the plane is using air resistance and acceleration is increasing while it goes up. Although you don’t know speed, still yes.

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Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m

const2013 [10]

Answer:

The answer to your question is vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

hmax = vo²sinФ/ 2g

Solve for vo²

vo² = 2ghmax / sinФ

Substitution

vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

vo² = 19.62(1.3) / 0.866

vo² = 25.51 / 0.866

vo² = 29.45

Result

vo = 5.43 m/s

5 0

3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$2M_2V_1^2=0.5\times M_2V_2^2$

$2V_1^2=0.5\times V_2^2$

$4V_1^2= V_2^2$

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

$2(V_1+9.0)^2=(2V_1+9.0)^2$

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

$(0.404V_1)=(3.72)$

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers

A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh

MrMuchimi

Answer:

W=700 Joule

Explanation:

Physics Work

Is the dot product of the force vector by the displacement vector

$W=\vec F \cdot \vec r$

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

$W=F.d$

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

$W=350\ N\ .\ 2\ m=700\ Joule$

4 0

3 years ago

A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material suc

notka56 [123]

Explanation:

A metal such as copper is a conductor because it provides a pathway for electric charges to move easily. A material such as rubber is an insulator because it resists the flow of electric charges. A material that partially conducts electric current is a semiconductor. These materials include group 3 and group 5 elements.

6 0

3 years ago

Read 2 more answers

A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c

Alekssandra [29.7K]

Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

6 0

2 years ago