A kayak on a lake is moving at a constant speed of 2.4 m/s in a direction 41.2° north of west. After 27 min, find the following.

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le

Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

$R = u² sin2θ/g$

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

$u² = \frac{Rg}{sin2\theta}$

$u^2 = \frac{9.1 x 9.80}{sin26}$

$u^2 = 113.17$

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

$v^2 = (v_osin\theta)^2 -2gh$

$h =\frac{(v_o^2sin\theta)^2}{2g}$

$h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}$

h = 1.10 m

7 0

2 years ago

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo

12345 [234]

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

$W = F\times d$

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute $W = 80.2\ J\ and\ d =25.0\ m$ in work done formula.

$80.2 = F\times 25$

$F=\frac{80.2}{25}$

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

3 0

2 years ago

A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.

As the acceleration of the object is constant, so

$a=\frac {v-u}{t}\cdots(i)$

Given that applied force, F=6.00 N,

From Newton's second law, we have

$F= m\times a$ ,

$\Rightarrow F=\frac {m(v-u)}{t}$ [from equation (i)]

$\Rightarrow Ft=m(v-u)$

$\Rightarrow Ft=mv-mu$

$\Rightarrow mv-mu=6\times 10$ [given that time, t=10 s and F=6 N]

$\Rightarrow mv-mu=60 kg \;m/s$

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

6 0

2 years ago

there are two slides at the park between which you are deciding. Both start at the height of 6 meter. One is short and Steve whi

kumpel [21]

It doesn't matter. If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
potential energy you had at the top, no matter what kind
of route you took getting down.
___________________________

The only way I can think of that it would make a difference
would be if the shallow slide were REALLY REALLY long,
and you didn't have anything to eat all the way down.
Then you might lose some weight while you're on the slide,
and your mass might be less at the bottom than it was at the
top. Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.

But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.

6 0

2 years ago

Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are

dolphi86 [110]

Answer:

a) P1+P2

Explanation:

The magnitude of their combined momentum is just the addition of each momentum, because in this case of inelastic collision, the kinetic energy of the two cars are both converted to some form of energy because the velocity of both cars becomes zero, i.e., V=0, making P = mv = 0, this means the magnitude of P1 + P2 = 0.

3 0

2 years ago