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Lostsunrise [7]
2 years ago
14

True or False: The basketball should be dribbled below the waist.

Physics
1 answer:
zimovet [89]2 years ago
7 0
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.
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John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
1. An electric iron has a rating of 750W, 220V. Calculate:
horrorfan [7]
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a). 6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span> For 7, realize that because they are in parallel, their voltages are the same. We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span> For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
6 0
3 years ago
already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0
3 years ago
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
Troyanec [42]

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between  loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

AC^2=AB^2+BC^2

AC^2=2.00^2+5.50^2

AC=\sqrt{(2.00^2+5.50^2)}

AC=5.85\ m

We need to calculate the path difference

Using formula of path difference

\Delta x=AC-BC

Put the value into the formula

\Delta x=5.85-5.50

\Delta x=0.35\ m

We need to calculate the lowest possible frequency of sound

Using formula of frequency

f=\dfrac{nv}{\Delta x}

Put the value into the formula

f=\dfrac{1\times340}{0.35}

f=971.4\ Hz

Hence, The lowest possible frequency of sound is 971.4 Hz.

8 0
3 years ago
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