All categories

5 days ago

6. Two blocks are released from rest at the same height. Block A slides down a steeper ramp than Block B. Both ramps are frictio

nless. The blocks reach the same final height indicated by the lower dashed line. Block B weighs more than Block A.
a. Is the work done by the gravitational force on Block A (greater/less than/equal to) the work done by the gravitational force on Block B? Explain your reasoning.

b. Is the speed of Block A (greater/less than/equal to) the speed of Block B? Explain your reasoning.

c. Is the momentum of Block A (greater/less than/equal to) the momentum of Block B? Explain your reasoning.

Physics

You might be interested in

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [3030]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

3 months ago

a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

inna [3103]

Definamos h como la distancia que hay desde el borde del pozo hasta la superficie del agua (en metros).

Consideremos la gravedad g como 9.8 m/s² y despreciemos la resistencia del aire.

La velocidad inicial vertical del guijarro es nula.
Ya que el guijarro impacta el agua tras 1.5 segundos, entonces:
h = 0.5 * (9.8 m/s²) * (1.5 s)² = 11.025 m

Resultado: 11.025 m

7 0

3 months ago

Read 2 more answers

Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc

ValentinkaMS [3465]

Answer:

$V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}$

Explanation:

The four wires are arranged in series: this setup indicates that the same current passes through them, while the total voltage from the battery, V0, equals the combined voltages across each resistor:

$V_0=V_1+V_2+V_3+V_4$