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7 days ago

Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13

.0 m/s. While running away at 3.00 m/s, Tex jumps onto the back of the bull to avoid being trampled. How fast does the bull run with Tex aboard?

Physics

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A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

$a=\frac{v^2}{R}$

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = $9.81\ m/s^2$

v = ?

R = 13.2 m

Por lo tanto,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

v = 11.4 m/s

8 0

3 months ago

A sinusoidal electromagnetic wave of frequency 6.10×1014hz travels in vacuum in the +x direction. the magnetic field is parallel

Yuliya22 [3333]

Part a) The connection between the electric field and the magnetic field in an electromagnetic wave is
$E=cB$
where
E signifies the strength of the electric field
B indicates the strength of the magnetic field
c represents the speed of light
Using the equation, we determine:
$E=cB=(3 \cdot 10^8 m/s)(5.80 \cdot 10^{-4} T)=1.74 \cdot 10^5 N/C$

Part b) The text does not clarify the orientation of the magnetic field on the y-axis: I speculate it points in the y+ direction.
The direction of the electric field can be established using the right-hand rule, which states:
- the index finger shows the direction of E
- the middle finger indicates the orientation of B
- the thumb reveals the propagation direction of the wave
Because the wave propagates in the x+ direction, and the magnetic field in the y+ direction, we conclude that the electric field direction (index finger) must be z-.

7 0

2 months ago

Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w

Keith_Richards [3271]

Solution:

$Em_{f}$ / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

K = ½ m v²

U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

Em₀ = K = ½ m v²

At its highest point

$Em_{f}$ = U = mg and

Now let's compute

Em₀ = ½ m 3.6²

Em₀ = m 6.48

$Em_{f}$ = m 9.8 × 0.2

$Em_{f}$ = m 1.96

Thus the energy lost is given by:

$Em_{f}$ / Em₀ = m 1.96 / m 6.48

$Em_{f}$ / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

8 0

2 months ago

Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnit

Yuliya22 [3333]

Answer:

The kinetic energy is higher for the first cart.

Explanation:

For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to

$(2kg)v = 10kg\: m/s$

resulting in

$v = 5m/s$ .

Consequently, the kinetic energy for the 3kg cart ends up as

$K.E. = \dfrac{1}{2}mv^2$

$= \dfrac{1}{2}(2kg)(5m/s)^2 = 25J$

$\boxed{K.E = 25J}$

indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.

3 0

3 months ago