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3 years ago

Explain Distance-Time Graphs and how to analyze it.

1 answer:

7 0

Distance-Time is what is used to show the VELOCITY of an object in a graph since Velocity=Distance traveled * Time taken ( V=S*T)
The Y-axis is the Velocity
The X-axis is the Time.

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The first physicist gets a second physicist to help. They both push on the crate, parallel to the surface of the incline, and it

Romashka-Z-Leto [24]

Answer:

F_aplied = fr

Explanation:

Newton's second law states that the force is proportional to the acceleration of the system, as in this case they indicate that the body moves at constant speed, the acceleration is zero, therefore

F_applied - fr = 0

F_aplied = fr

therefore the force applied by people is equal to the friction force

3 0

3 years ago

Helo am desperate answer quick will give brainliest

Anika [276]

Answer:

Pitch. The sound an article makes changes relying upon how quick it is vibrating. At the point when an item vibrates rapidly, sharp sounds are heard. Low-pitched sounds come from things that vibrate all the more gradually.

Explanation:

7 0

3 years ago

Read 2 more answers

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

4 years ago

Annie has a soccer ball and a cake ball she kicks each ball with the same Force the soccer ball accelerates at 3 m per second sq

lana [24]

Answer: The acceleration will be greater in the kickball than the soccer ball since the kick ball has a smaller mass and provided they are acted upon with equal force magnitude.

Explanation:

We can prove this by substituting with numerical values.

Say the force she kicked both balls with was 30 N. The mass of the soccer ball is 10kg and the mass of the kickball is 6kg.

Applying Newtons Second Law, F = ma

For the soccer ball:

30 = 10a
a = 30/10
a = 3

For the kickball:

30 = 6a
a = 30/6
a = 5

Hence the kickball will travel with a greater acceleration given the same force is acted upon and has a smaller mass than the soccer ball

Hope this helps, brainliest would be appreciated :)

4 0

3 years ago

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)