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3 years ago

A hiker climbs a hill 400 m high. If he weighs 50 kg

Physics

2 answers:

Zarrin [17]3 years ago

5 0

Answer:

20 is Solvent this is the answer u divide them

user100 [1]3 years ago

3 0

The work done on an object is directly proportional to the force on the object, the displacement of the object, and the cosine of the angle between the force and the displacement:

W = Fd cos 0

Energy is the ability to do work.

Work is a scalar quantity, which is usually measured in newton-meters or joules.

Work is 0 when the force is perpendicular to an object's displacement or when the object isn’t displaced.

You might be interested in

How far away is the sun?

Serhud [2]

It changes slightly during a year. The average is about 93 million miles.

5 0

3 years ago

A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista

omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

7 0

2 years ago

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

3 years ago

Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocit

ser-zykov [4K]

The reaction time of Boris is t(r), so before that, Boris will not have jumped. Thus, H(b)(t) = 0
The vertical displacement will simply be
D(t) = H(a)(t)

7 0

3 years ago

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km

$tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2$

$\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees$ north or west or (180 - 72.65) = 107.35 degrees north of east

3 0

3 years ago