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3 years ago

How could the action force of a canoe moving through water be increased?.

1 answer:

4 0

There are several actions that can be used to increase the action force of a canoe that is moving through water. One can increase the rate of striking or strike harder than before with a bigger paddle. Using a lighter canoe can also be helpful for increasing the speed. I hope the answer comes to your help.

You might be interested in

What energy transformations take place when you eat a taco and then go for a jog

sesenic [268]

Answer:

Chemical to kinetic and thermal.

Explanation:

You would eat the food (chemical) than you would jog and move around (kinetic). While running your body would also give off heat (thermal).Than your body would sweat to cool itself down.

6 0

3 years ago

A football punter accelerates a .55 kg football

Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the football

$\Delta t=0.25 s$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.55 kg is the mass of the football

u = 0 is the initial velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we find the force exerted on the ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N$

5 0

3 years ago

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horiz

igomit [66]

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^{-6} T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

3 0

3 years ago

Imagine that a bowling ball needs to be lifted 1.5 meters, and its gravitational potential energy is 90 joules. How much does th

Vikki [24]

Answer: 6.12 kg

Explanation:

Since Mass of ball = ? (let the unknown value be Z)

Acceleration due to gravity, g= 9.8m/s^2

Height, h = 1.5 metres

Gravitational potential energy GPE = 90J

Gravitational potential energy depends on the weight of the ball, the action of gravity and height.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

90J = Z x 9.8m/s^2 x 1.5m

90 = Z x 14.7

Z = 90/14.7

Z = 6.12 kg

Thus, the bowling ball weigh 6.12 kilograms

8 0

3 years ago

A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails

harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

$y=h+ut+\frac{1}{2}gt^2$

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

$y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m$

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

$v^2-u^2 = 2ad$

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

$v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s$

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

$a=\frac{v-u}{t}$

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

$t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s$

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0

2 years ago