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gtnhenbr [62]
3 years ago
11

Ask Your Teacher The height h in feet reached by a dolphin t seconds after breaking the surface of the water is given by h = −16

t2 + 32t. How long will it take the dolphin to jump out of the water and touch the trainer's hand when h = 16?
Physics
1 answer:
kirza4 [7]3 years ago
3 0

Answer:

1 second

Explanation:

h = −16t² + 32t

When, h = 16

16 = −16t² + 32t

Divide each of the numbers by 16

1 = -1t² + 2t

Rearrange the equation

1t²-2t+1 = 0

Solving by the quadratic formula, we get

t=\frac{-(-2)\pm \sqrt{(-2)^2-4\times 1\times 1}}{2\times 1}\\\Rightarrow t=\frac{2\pm 0}{2}\\\Rightarrow t=1\ s

So, time taken by the dolphin to jump out of the water and touch the trainer's hand is 1 second.

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What does the negative sign in F = –kx mean?
VashaNatasha [74]

It says that the force of a spring is always opposite to the direction in which you stretch it or compress it. (the direction of 'x')

In other words, a spring that's disturbed always tries to put itself back to it's normal length, where x would be zero.

8 0
3 years ago
We move a square loop of wire 2.3 cm on a side into a region of uniform magnetic field of 1.5 T with the plane of the loop perpe
vova2212 [387]

Answer:

Force will be F=8.33\times 10^{-4}N

Explanation:

We have given side of wire d = 2.3 cm = 0.023 m

Magnetic field B = 1.5 T

Velocity v = 2.1 m/sec

Resistance R = 3 ohm

We know that current is given by i=\frac{Bvd}{R}=\frac{1.5\times 2.1\times 0.023}{3}=0.02415A

Power is given by P=i^2R=0.02415^2\times 3=0.00174watt

We know that power is also given by P=Fv

So 0.00174=F\times 2.1

F=8.33\times 10^{-4}N

4 0
3 years ago
The mass of a sports car is 1000 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.260 and the frontal
alisha [4.7K]

Answer:

The initial acceleration is 0,221 m/s^2.

Explanation:

Fcar = Fairfriction\\m*a = 0.5*A*c*density*v^2\\a = 0.5*A*c*density*v^2 / 1000 kg = 0.5*2.10m^2*0.260*1.295kg/m^3*(25 m/s)^2 / 1000 kg = 0,221 m/s^2

8 0
3 years ago
Read 2 more answers
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of lengt
damaskus [11]

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure,  41.0  PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be  14.6959  P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂    ⇒    V₂ = P₁*V₁ / P₂

where  

1  and  2  are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

P₁ = 14.6959  P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒  V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A  = (13.2985 in³) / (3.00 in²)

⇒  h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

4 0
3 years ago
Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.
Sati [7]

Answer:

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836, z^{\frac{1}{3} }= -1.967 - i\cdot 2.265, z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]

Where angles are measured in radians and k represents an integer between 0 and n - 1.

The magnitude of the complex number is 27 and the equivalent angular value is 1.817\pi. The set of cubic roots are, respectively:

k = 0

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836

k = 1

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]

z^{\frac{1}{3} }= -1.967 - i\cdot 2.265

k = 2

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]

z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

5 0
3 years ago
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