Ask question

Ask question

All categories

2 years ago

6

On a straight road (taken to be in the +x direction) you drive for an hour at 50 km per hour, then quickly speed up to 90 km per

hour and drive for an additional two hours.Required:a. How far do you go?b. What is your average x component of velocity?c. Why isn't vavg,x equal to the arithmetic average of your initial and final values of vx, (50+90)/2 = 70 km per hour?

1 answer:

Rufina [12.5K]2 years ago

5 0

Answer:

A.) 230 km

B.) 76.67 km/h

Explanation:

Given that On a straight road (taken to be in the +x direction) you drive for an hour at 50 km per hour, then quickly speed up to 90 km per hour and drive for an additional two hours.

A.) How far do you go?

When driving for an hour, the distance covered will be

Distance = speed × time

Distance = 50 × 1 = 50 km

When driving for additional 2 hours, the distance covered will be

Distance = 90 × 2 = 180 km

Total distance = 180 + 50 = 230 km

b. What is your average x component of velocity?

Average Velocity = total distance/ total time

Average velocity = 230/3

Average velocity = 76.67 km/h

c. Why isn't vavg,x equal to the arithmetic average of your initial and final values of vx, (50+90)/2 = 70 km per hour

They are not equal because of the displacement is the same as distance travelled.

You might be interested in

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

notsponge [240]

Answer:

$Q = \frac{0.068}{E}$

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

$QE = F_x$

$mg = F_y$

since string makes 30 degree angle with the vertical so we will have

$tan\theta = \frac{F_x}{F_y}$

$tan30 = \frac{QE}{mg}$

$Q = \frac{mg}{E}tan30$

$Q = \frac{0.012\times 9.81}{E} tan30$

$Q = \frac{0.068}{E}$

where E = electric field intensity

5 0

2 years ago

The heaviest wild lion ever measured had a mass of 313 kg. Suppose this lion is walking by a lake when it sees an empty boat flo

11Alexandr11 [23.1K]

Answer:

The kinetic energy dissipated is 3286.5 J

Explanation:

K.E before collision = 1/2m1v1^2 = 1/2×313×6^2 = 5634 J

K.E after collision = 1/2(m1+m2)v2^2

From the law of conservation of momentum:

m1+m2 = m1v1/v2 = 313×6/2.5 = 751.2 kg

K.E after collision = 1/2×751.2×2.5^2 = 2347.5 J

K.E dissipated = 5634 J - 2347.5 J = 3286.5 J

3 0

2 years ago

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider dives to a height of 278 mete

Verdich [7]

Answer:

<em>The glider's new speed is 68.90 m/s</em>

Explanation:

<u>Principle Of Conservation Of Mechanical Energy</u>

The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

$\displaystyle K=\frac{1}{2}mv^2$

The gravitational potential energy when it's at a height h from the zero reference is

$U=mgh$

The total mechanical energy is

$M=K+U$

$\displaystyle M=\frac{1}{2}mv^2+mgh$

The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.

The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

$\displaystyle M_1=\frac{1}{2}m(45.2)v_o^2+m(9.8)(416)$

Operating in terms of m

$\displaystyle M_1=1021.52m+4076.8m$

$\displaystyle M_1=5098.32m$

Then we know the glider dives to 278 meters and we need to know their final speed, let's call it $v_f$ . The final mechanical energy is

$\displaystyle M_2=\frac{1}{2}mv_f^2+m(9.8)(278)$

Operating and factoring

$\displaystyle M_2=m(\frac{1}{2}v_f^2+2724.4)$

Both mechanical energies must be the same, so

$\displaystyle m(\frac{1}{2}v_f^2+2724.4)=5098.32m$

Simplifying by m and rearranging

$\displaystyle \frac{v_f^2}{2}=5098.32-2724.4$

Computing

$v_f=\sqrt{4747.84}=68.90\ m/s$

The glider's new speed is 68.90 m/s

8 0

2 years ago

49. A vertically hung 0.50-meter- long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight a

Natali5045456 [20]

Answer:

$K=120$

Explanation:

From the question we are told that

Length of spring $l_1=0.5m$

Length of stretched $l_s=1m$

Potential energy of spring $E=15J$

Generally equation for energy stored is mathematically given as

$U=1/2K \triangle x^2$

$K=\frac{2U}{\triangle x^2}$

$K=\frac{2*15}{ 0.5^2}$

Therefore value of the spring constant in N/m? is given as

$K=120$

4 0

1 year ago

An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

tamaranim1 [39]

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

5 0

2 years ago