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3 years ago

13

A ball is thrown with an initial speed of 10. Meters per second. At what angle above the horizontal should the ball be thrown to

reach the greatest height?

2 answers:

maria [59]3 years ago

6 0

The initial speed doesn’t matter. To reach the maximum height, you need to launch straight up ... 90 degrees above horizontal.

Jlenok [28]3 years ago

5 0

Answer:

90 degree

Explanation:

The height for the projectile motion is given by

H = u^2 Sin^2 theta/ 2g

Height is maximum when sin theta is maximum.

Sin theta is maximum when it's value is 1.

So the value of theta is 90 degree.

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A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

$W_f = F_fs = 15*0.5 = 7.5 J$

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

$E_p = mgh = 0.1*10*1.5 = 1.5 J$

Therefore, as elastic energy is converted to potential energy and work of friction:

$E_e = W_f + E_p$

$kx^2/2 = 7.5 + 1.5 = 9 J$

$k = 9*2/x^2 = 18/0.05^2 = 7200 N/m$

6 0

3 years ago

Four forces act on a hot-air balloon, as shown

dolphi86 [110]

The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

<em>Upward force = 514 N</em>
<em>Downward force = 267 N</em>
<em>Eastward force = 678 N</em>
<em>Westward force = 397 N</em>

The net vertical force on the balloon is calculated as follows;

$F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N$

The net horizontal force on the balloon is calculated as follows;

$F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N$

The magnitude of the resultant force on the balloon is calculated as follows;

$F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N$

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

Learn more here:brainly.com/question/4404327

5 0

2 years ago

A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?

ser-zykov [4K]

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

${ \rm{ {v}^{2} = {u}^{2} + 2gs}}$

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

${ \rm{0 {}^{2} = {2}^{2} - (2 \times 9.8 \times s)}} \\ \\ { \rm{4 = 19.6s}} \\ \\ { \rm{s = 0.204 \: m}}$

4 0

2 years ago

Read 2 more answers

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

<u>Second stone</u>

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago

Pls answerrrrr thisssss

Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0

3 years ago