Ask question

Ask question

All categories

3 years ago

10

A freight train rolls along a track with considerable momentum. If it rolls at the same speed but hastwice as much mass, its mom

entum isa. zero.b. doubled.c. quadrupled.d. unchanged.

1 answer:

Margaret [11]3 years ago

8 0

Its Momentum Would Be Doubled

You might be interested in

In 1935, a French destroyer, La Terrible, attained one of the fastest speeds for any standard warship. Suppose it took 3.0 min a

REY [17]

Answer:

Explanation:

From newton's equation of motion of uniform acceleration

v = u + at

where v is final velocity , u is initial velocity , a is acceleration and time is t .

putting the values

v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )

= 90 m /s

So , final velocity is 90 m /s .

7 0

4 years ago

A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure

klasskru [66]

Answer:

$3x_0$

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

$d=v_0 t = x_0$

in the second case, the horizontal velocity is increased to

$v_x' = 3v_0$

And so the new distance travelled will be

$d' = v_x' t = 3 v_0 t = 3 x_0$

So, the distance increases linearly with the horizontal velocity.

5 0

4 years ago

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

4 years ago

A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert

strojnjashka [21]

Answer:

$v=10m/sec$

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

$F_N+Fg$

where

$F_N =forceof the normal$

$Fg= force due to gravity$

Generally the net force is given to be $FC(force towards center)$

$F_C =F_N + Fg$

$F_N =F_C -Fg$

$F_N=F_C-F_g$

$F_N=\frac{mv^2}{R} -mg$

Mathematical we can now derive V

$m_g + \frac{8m}{4}= \frac{mv^2}{8}$

$\frac{5mg}{4} =\frac{mv^2}{8}$

$v^2 =\frac{40*10}{4}$

$v=10m/sec$

Therefore the speed of the roller coaster is given ton be $v=10m/sec$

5 0

3 years ago

An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti

Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
So, we can write the following expression for Fg:

$F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)$

b)

The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

$F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N (2)$

c)

According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

$a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2 (3)$

3 0

3 years ago