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iris [78.8K]
3 years ago
13

A man weighing 720 N and a woman weighing 500 N have the same momentum. What is the ratio of the man's kinetic energy Km to that

of the woman Kw?
Physics
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662Explanation:

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Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
A car is traveling at 24.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant accelerat
Elan Coil [88]

Answer:

The answer to your question is : vf = 15.18 m/s

Explanation:

Data

vo = 24 m/s

d = 120 m

vf = ? when d = 60.0 m

Formula

              vf² = vo²  + 2ad

For d =100m

            a = (vf² - vo²) / 2d

            a = (0 -24²) / 2(100)

            a = -576/200

             a = 2.88 m/s²

Now, when d = 60

        vf² = (24)² - 2(2.88)(60)

        vf² = 576 - 345.6

        vf² = 230.4

        vf = 15.18 m/s

4 0
3 years ago
Read 2 more answers
Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t
nlexa [21]

Answer:

Explanation:

Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.

Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

Tsinθ = F ( F is force of repulsion between two charges sphere)

Dividing the two equations

Tanθ = F / mg

tan17 = F / (7.1 x 10⁻³ x 9.8)

F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17  = .41 m )

21.27 x 10⁻³  = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

no of electrons required  = q / charge on a single electron

= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .39375 x 10¹³

3.9375 x 10¹² .

4 0
3 years ago
A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou
tamaranim1 [39]

We have that the amount of the charge q is

q=1.8*10^{-7}

From the Question we are told that

Force F=7.89 x*10^{-7}

Velocity V=2090m/s

Angle \theta=29.4

Magnetic field B=4.23 * 10^{-3} T

Generally, the equation for Force F is mathematically given by

F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}

q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}

q=1.8*10^{-7}

In conclusion

The amount of the charge q is

q=1.8*10^{-7}

For more information on this visit

brainly.com/question/1470439

6 0
3 years ago
Read 2 more answers
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
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