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3 years ago

Why doesn't the force of gravity change the speed of a bowling ball as it rolls along a bowling lane?

1 answer:

5 0

Because the gravitational force, which points downward, is perfectly balanced by the normal reaction of the floor of the bowling lane, which points upward. The two forces are equal in magnitude, so the net force acting vertically on the bowling ball is zero, therefore there is no acceleration along this direction. Moreover, since the ball is moving in the horizontal direction, the gravitational force has no component along this direction, so it does not change the velocity of the ball.

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Dogs are able to hear much higher frequencies then humans are capable of detecting. For this reason the crystals that are inaudi

AVprozaik [17]

Answer:

Dogs are able to hear much higher frequencies than humans are capable of detecting. For this reason, dog whistles that are inaudible to the human ear can be heard easily by a dog. If a dog whistle has a frequency of 30,000 Hz, what is the wavelength of the sound emitted?

3 0

3 years ago

Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as

Lesechka [4]

We are given:

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

__________________________________________________________

Momentum of Tarzan before swinging:

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

Momentum = 560 kg m/s

__________________________________________________________

Speed of Tarzan after grabbing the bananas:

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v [where v is the velocity of Tarzan]

v = 5.9 m/s

__________________________________________________________

<h3>Finding the Initial and Final KE and PE:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

Initial and Final KE:

We know that KE = 1/2*(mv²)

Initial KE:

Initial KE = 1/2*(mv²) [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

Final KE:

Final KE = 1/2*(mv²)

[where v is the velocity at the maximum point of the swing]

Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

Initial and Final PE:

We know that:

PE = mgh

[where g is the acceleration due to gravity and h is the height]

Initial PE:

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

Final PE:

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3>Finding the maximum height Tarzan will reach:</h3>

Here, KE = Kinetic Energy and PE = Potential Energy

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

1653.5 + 0 = 0 + 950h

1653.5 = 950h

h = 1653.5/950 [dividing both sides by 950]

h = 1.74 m

Therefore, the maximum height reached by Tarzan is 1.74 m

but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey

6 0

2 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

3 years ago

When water freezes, its volume increases by 9.05% (that is, equation). What force per unit area is water capable of exerting on

PIT_PIT [208]

Answer:

1.991 × 10^(8) N/m²

Explanation:

We are told that its volume increases by 9.05%.

Thus; (ΔV/V_o) = 9.05% = 0.0905

To find the force per unit area which is also pressure, we will use bulk modulus formula;

B = Δp(V_o/ΔV)

Making Δp the subject gives;

Δp = B(ΔV/V_o)

Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²

Thus;

Δp = 2.2 × 10^(9)[0.0905]

Δp = 1.991 × 10^(8) N/m²

3 0

2 years ago

Which is an example of qualitative data?

LiRa [457]

The answer is C)the rating that the golfers give Callaway clubs

6 0

2 years ago