Ask question

Ask question

All categories

anzhelika [568]

3 years ago

12

John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

he distance and displacement?

1 answer:

KIM [24]3 years ago

7 0

The distance is 30 km and the displacement is 22.4 km North East

You might be interested in

What vectors make up R

Tju [1.3M]

Strange as it may seem, I don't think ANY of those choices will work. Anybody else ?

8 0

3 years ago

A small economy car (low mass) and a limousine (high mass) are pushed from rest across a parking lot, equal distances with equal

Studentka2010 [4]

Answer:

The car that receives more kinetic energy is the small economy car.

Explanation:

K.E = 0.5*mv²

Where;

K.E is the kinetic Energy

M is the mass of an object

V is the velocity of the moving object

But F = m(v/t), from Newton's second law of motion

If equal forces were applied to the two cars, then the velocity of each car will be calculated as follows.

v = (Ft/m)

v² = (Ft/m)²

Substitute in the value of v² into Kinetic energy equation

K.E = 0.5*mv²

K.E = 0.5*m(Ft/m)² = (0.5*F²t²)/m

Also assuming equal distance, equal force and assuming equal time for both cars.

The above equation will reduce to, K.E = k/m

Where k = 0.5*F²t², which is equal in both cars.

Thus, Kinetic energy will depend only on the mass of each car.

From the above expression, Kinetic Energy received by each car is inversely proportional to the mass of the car.

A small economy car (low mass) will receive more kinetic energy while a limousine (high mass) car will receive less kinetic energy.

Therefore, the car that receives more kinetic energy is the small economy car.

6 0

3 years ago

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

or

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago

Which of the following equations can be used to directly calculate an object's impulse?

Iteru [2.4K]

I believe the answer is B

7 0

3 years ago

A historical society is testing an old cannon. They

sveta [45]

Answer:

26.2 m/s

Explanation:

We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is

$u_y = u sin \theta$ (1)

where u is the initial speed and $\theta = 45.0^{\circ}$ is the angle of projection.

We can therefore use the following suvat equation for the vertical motion of the ball:

$v_y = u_y + at$

where $v_y$ is the vertical velocity at time t, and $a=g=-9.8 m/s^2$ is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:

$t=\frac{3.78}{2}=1.89 s$

And at the maximum height, the vertical velocity is zero:

$v_y=0$

Substituting these values, we find the initial vertical velocity:

$u_y = v_y - at = 0-(-9.8)(1.89)=18.5 m/s$

And using eq.(1) we now find the initial speed:

$u=\frac{u_y}{sin \theta}=\frac{18.5}{sin 45.0^{\circ}}=26.2 m/s$

4 0

3 years ago