Ask question

Ask question

All categories

3 years ago

12

A fielder tosses a 0.15 kg baseball at 26 m/s at a 36 ∘ angle to the horizontal.Part AWhat is the ball's kinetic energy at the s

tart of its motion?Express your answer with the appropriate units.Part BWhat is the kinetic energy at the highest point of its arc?Express your answer with the appropriate units.

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

A. 50.7 J

B. 33.18 J

Explanation:

At the start of the motion the kinetic energy of the ball would be

$E_k = \frac{mv^2}{2} = \frac{0.15*26^2}{2} = 50.7 J$

When the ball gets to the highest point, vertical velocity must be 0, only horizontal velocity contributes to the kinetic energy. Since air resistant can be neglected, horizontal energy is preserved since the start

$v_h = vcos(36^o) = 26*0.81 = 21.03 m/s$

So the kinetic energy at this point is

$E = \frac{mv_h^2}{2} = \frac{0.15*21.03^2}{2} = 33.18J$

You might be interested in

Which stage of star development comes right after the long stable phase in the life cycle

Bond [772]

I believe It's either a supernova or a planetary nebula.

6 0

3 years ago

Compare the kinematics equation governing the motion y =vf+1/2 gf' and the format of the fit equation that you just made y At +B

lidiya [134]

Answer:

please find solution attached

Explanation:

3 0

3 years ago

Read 2 more answers

If two bodies weighing 10 kg and 20 kg respectively are standing at the

kvv77 [185]

Answer:

potential energy = mgh , where m is mass of body , g is acceleration due to gravity and h is given height

since both of the body are at same height above the ground , so the body having greater mass will have more potential energy

that is 20 kg body will have higher P.E

Explanation:

7 0

3 years ago

A 12cm candle is placed 6cm from a converging lens with a focal length of 15cm. What is the height of the image of the candle? S

amm1812

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

v is image distance

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm$

If h' is the height of the image. Magnification is given by :

$m=\dfrac{h'}{h}=\dfrac{v}{u}$

$h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm$

So, the height of the image of the candle is 20 cm.

3 0

3 years ago

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )

$x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)$

$x_{C} = 128.995\,m$

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

0 0

3 years ago