Which planet is most likely to have acid rain? A. Mercury B. Venus C. Mars D. Uranus

Ocean currents near the equator are primarily: eastward westward north and south clockwise

sladkih [1.3K]

Equatorial currents are primarily westward. This is because the dominant current in the northern hemisphere has a clockwise direction, while the southern hemisphere has a counterclockwise direction. When these two currents meet at the equator, a common westward current exists.

3 0

4 years ago

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

3 years ago

Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul

Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

$v_1=\sqrt{2gh}$

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

$J=m(v_2-v_1)$

$J=2m\sqrt{2gh}$

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

7 0

4 years ago

How to do this, i'm completely lost

vaieri [72.5K]

There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°))

T = 1501,6 N

8 0

4 years ago

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k

Brrunno [24]

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$