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3 years ago

11

A young boy is dragging his 10 kg stuffed animal by the tail at a constant velocity of 4 m/s. The coefficient of friction betwee

n the dog and the ground is 0.4.

1 answer:

Daniel [21]3 years ago

7 0

Answer:

if you multiply the coefficient to the denominator you will find that the percentage of increase is equal to your ending point

Explanation:

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If themass is 50kg, what weight of water is to be displaced to float on water? why

ludmilkaskok [199]

Answer:if youre looking for the weight of the thermas in genral it should be 500n

Explanation:using the formula w=mg

w=500x10

giving us 500 newtons which is the weight.

5 0

1 year ago

Which of the following is the closest to the scientific fact

Finger [1]

B - A theory seems to be the closest

7 0

3 years ago

A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp

raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination $\theta =11.1^{\circ}$

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction $\mu _k=0.39$

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using $v^2-u^2=2as$

Final velocity v=0

$0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s$

$s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}$

s=0.68 m

5 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

Helga [31]

Answer: $3.23 m/s^2$

Explanation:

Given

$\mu_s =0.330$

Frictional Force is balanced by force due to car acceleration

Frictional force $F_s$

$F_s=ma_{min}$

$\mu_sN=ma_{min}$

$\mu_s\cdot mg=ma_{min}$

$a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2$

6 0

3 years ago

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0

3 years ago