Which item is heavier, 50 kg of cotton or 50 kg of iron?

QUICK! If i have a 95.6% in my as my grade in that class what is 30% off of that grade?

cluponka [151]

66.92 will be your grade

6 0

3 years ago

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

levacccp [35]

In component form, the displacement vectors become

• 350 m [S] ==> (0, -350) m

• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

5 0

2 years ago

The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an

yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$

$\frac{1}{f}=0.028$

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

$p=\frac{1}{f}$

Here, p is the power of the lens.

Put f= 35.71 cm.

$p=\frac{1}{35.71}$

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0

2 years ago

You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o

Serhud [2]

Answer:

The value is $h = 13.2 \ m$

Explanation:

From the question we are told that

The speed of the rope with hook is $u = 24 .1 \ m/s$

The angle is $\theta = 65.0^o$

The speed at which it hits top of the wall is $v = 16.3 m/s$

Generally from kinematic equation we have that

$v_y^2 = u_y ^2 + * 2 (-g)* h$

Here h is the height of the wall so

$[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h$

=> $h = 13.2 \ m$

4 0

2 years ago

Which statement is true regarding flexibility? A. joint's range of motion cannot be maintained with age. b.A joint's range of mo

Alex777 [14]

<span>the statement that is true regarding flexibility is : b. a joint's range of motion will be lost if the joint is not used regularly. Our body is like a machine. If we not constantly heat it up, our body will be more prone to injury. We can see that the old people who lived within the tribe in the middle of the mountain are far stronger than the one who lived in the city.</span>

3 0

3 years ago

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