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EleoNora [17]
3 years ago
5

A rock is thrown upward with a velocity of 23 meters per second from the top of a 25 meter high cliff, and it misses the cliff o

n the way back down. when will the rock be 11 meters from the water, below
Physics
1 answer:
Nana76 [90]3 years ago
8 0
Vf=vi+at
-23m/s=23m/s-9.8m/s^2(t)
t1 from throwing up back to starting position t1=4.693s
Delta(x)=vi(t)+.5a(t^2)
25-11=14
-14m=-23m/s(t)-4.8m/s^s(t^2)
t2 is from 25m to 11 m t2=.546s
Total time =5.239s
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MATHPHYS PLEASE HELP
solmaris [256]

Explanation:

Momentum is mass times speed.

p = mv

a) p = (1500 kg) (25.0 m/s) = 37,500 kg m/s

b) p = (40,000 kg) (1.00 m/s) = 40,000 kg m/s

The truck has more linear momentum.

Momentum in the y direction:

pᵧ = (1500 kg) (25.0 m/s) = 37,500 kg m/s

Momentum in the x direction:

pₓ = (1500 kg) (15.0 m/s) = 22,500 kg m/s

Total linear momentum:

p² = pₓ² + pᵧ²

p² = (22,500 kg m/s)² + (37,500 kg m/s)²

p = 43,700 kg m/s

3 0
3 years ago
Read 2 more answers
A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force
SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, \theta=30^{\circ}

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

7 0
3 years ago
The lowest point in Death Valley is 85 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m.What is th
mario62 [17]

Answer:

\Delta E=2.87\times 10^6\ J

Explanation:

It is given that,

Depth of Death valley is 85 m below sea level, h_i=-85\ m

The summit of nearby Mt. Whitney has an elevation of 4420 m, h_f=4420\ m

Mass of the hiker, m = 65 kg

We need to find the change in potential energy. It is given by :

\Delta E=mg(h_f-h_i)

\Delta E=65\times 9.8(4420-(-85))

\Delta E=2869685\ J

or

\Delta E=2.87\times 10^6\ J

So, the change in potential energy of the hiker is 2.87\times 10^6\ J. Hence, this is the required solution.

5 0
3 years ago
A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
3 years ago
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ANEK [815]

Hi!


The correct answer would be: the width of I-bands


The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).


<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>


<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.


Hope this helps!

8 0
3 years ago
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