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fiasKO [112]
3 years ago
7

I’ll name u BRAINLIEST if u get this right

Physics
1 answer:
nekit [7.7K]3 years ago
7 0
V = m1 u1 - m2 u2 / m1
v = 0.01 * 500 - 2 * 1.4 / 0.01
v = 220 m/s
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1. Assume the acceleration of gravity is 2 10 m/sec downwards. A cannonball is fired at ground level. If the cannon ball rises t
Ray Of Light [21]

Answer:

A. 50 m/s

Explanation:

So, we know that at the highest point of the cannonball’s path the y-component of the velocity is 0 since the projectile is not moving in that axis momentarily. We can set the following equation at that instant:

v_y =v_0y^2 +2gh \rightarrow 0 = v_0y^2 +2gh \rightarrow v_0y^2 =2gh

v_0y^2 =-2*(-10 m/s^2)*80 m=1600 m^2/s^2

v_0y =\sqrt(1600 m^2/s^2) = 40 m/s

Now that we found the y-component of the cannonball’s initial velocity, we can also find how much time it takes the projectile to reach the maximum height point, using the following equation:

v_y = v_0_y +a*t

We already stablished that in this point vy = 0. In that case:

v_0_y = -g*t \rightarrow 40 m/s=-(-10m/s^2)*t \rightarrow t=(40 m/s)/(10m/s^2) = 4s

In the cannonball’s path, the maximum height point is in the middle point of the total x displacement. In other words, the projectile takes about 2t to touch the ground. That means that for the projectile to reach x=240 m, it requires t = 2*4s = 8 s. now, we proceed and calculate the x-component of the initial velocity, using the following equation:

x = x_0 + v_0x+1/2*a_x*t^2

The projectile started at the point of reference, so we consider x0 = 0. Also, the velocity of the projectile maintains constant along the x-axis, so we consider ax=0 as well. Therefore, we have that:

x = v_0x*t \rightarrow 240 m = v_0x*(8s)

v_0x = (240 m)/(8s) = 30 m/s

The magnitude of the initial velocity can be found using the Pythagorean Theorem

v_0 = \sqrt{(30 m/s)^2 +(40 m/s)^2}= 50 m/s

So, the cannonball must be thrown with an initial velocity of 50 m/s to reach an height of 80 m and a displacement on the x-axis of 240 m

7 0
3 years ago
A 5 kg box is lying at rest on a frictionless surface. If you push it with a constant net force of 10 N, how far will it travel
Contact [7]

The distance traveled by the box over the first 4 s  is 16 m.

To solve the problem above,  First, we apply newton's Fundamental equation of force.

<h3 /><h3>Newton's fundamental equation of force:</h3>

F = ma...................... Equation 1

Where

  • F = Force
  • m = mass of the box
  • a = acceleration

make "a " the subject of the equation

a = F/m...................... Equation 2

From the question,

Given:

  • F = 10 N
  • m = 5 kg

Substitute these values into equation 2

a = 10/5

a = 2 m/s²

Finally, we use the equation of motion to calculate the distance it will travel over the given period of time.

<h3><u>Equation of motion</u></h3>

s = ut+at²/2........................... Equation 3

Where:

  • s = total distance traveled
  • t = time
  • u = initial velocity

Given:

  • t = 4 s
  • u = 0 m/s (at rest)
  • a = 2 m/s²

Substitute these values into equation 3

  • s = 0(4)+(2×4²)/2
  • s = 0+32/2
  • s = 16 m

Hence, the distance traveled by the box over the first 4 s  is 16 m

Learn more about Newton's fundamental equation here: brainly.com/question/13370981

6 0
2 years ago
A liquid is flowing through a horizontal pipe whose radius is m. The pipe bends straight upward through a height of 10.6 m and j
zaharov [31]

Answer:

Volume flow rate = 1.81 * 10^{-2} meter cube per second

Explanation:

As we know that the

Pressure at the two ends would be the same along with volume of flow.

i.e

P_1 = P_2

and

A_1 V_1 = A_2 V_2

Re arranging the file, we get -

V_1 = \frac{A_2 V_2}{A_1}

The flow equation is

\frac{1}{2}\rho * V_1^2 = \frac{1}{2}\rho * V_2^2 + \rho * g * h\\

Substituting the value of V_1 in above equation, we get -

V_2 = \sqrt{\frac{2gh}{(\frac{A_2}{A_1})^2-1} }

Substituting the given values in above equation we get

V_2 = \sqrt{\frac{2*9.8*10.6}{(\frac{\pi 0.04^2}{\pi 0.02^2} )^2 -1} }\\ V_2 = 3.61 m

Volume flow rate

Q_2 = A_2 V_2\\= \pi r_2^2V_2^2\\= 3.14 * 0.04^2 * 3.61 \\= 1.81 * 10^{-2}

7 0
3 years ago
When you ride a bicycle, in what direction is the angular velocity of the wheels? When you ride a bicycle, in what direction is
Hitman42 [59]

Complete question is;

When you ride a bicycle, in what direction is the angular velocity of the wheels? A) to your left B) to your right C) forwards D) backwards

Answer:

Option A - to your left

Explanation:

While an object rotates, each particle will have a different velocity:

the 'Speed' component will vary with radius while the 'Direction' component will vary with angle.

All of the velocity vectors are aligned in the same plane.

We can be solve this by choosing a single vector normal to ALL of the possible velocity vectors of the rotating object in that plane.

This convention used is known as "Right-hand rule". The angular velocity vector points along the wheel's axle. For instance, if you Imagine wrapping your right hand around the axle so that your fingers point in the direction of rotation, with your thumb sticking out. You will notice that your thumb points to the left.

Thus;

By right-hand rule, a wheel rotating on a forward - moving bicycle has an angular velocity vector pointing to the rider's left.

So, option A is the correct answer

3 0
4 years ago
What must be true about a surface in order for diffuse reflection to occur?
balu736 [363]

Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

4 0
3 years ago
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