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3 years ago

For an extended object, the weight force can be considered to act at which point?

Physics

1 answer:

JulsSmile [24]3 years ago

5 0

Answer:

At the center of the object

At the end of the object farthest away from the ground

At the center of gravity of the object

At end of the object closest to the ground

Explanation:

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Distinguish between resultant and equilibrant forces<br><br><br>

SIZIF [17.4K]

Answer:

resultant is a single force that can replace the of a number of forces , equililbrant is a force that is exactly opposite to resultant

Explanation:

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3 years ago

1. Gwen exerts a 16 N horizontal force as she pulls a 32N sled across a cement

PSYCHO15rus [73]

Answer:

3.6 N Kinetic friction

Explanation:

3 ) Ff = uk * Fn = 3.6 N

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2 years ago

Why do our minds fill in the gaps for information we don't know? (Hint: remember what you learned about Gestalt psychology.)

shutvik [7]

the most appropriate answer is A !! our mind automatically connects everything and thus make a story that we don't even familiar with in actual !!

7 0

3 years ago

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

3 years ago

Wildlife biologists are attempting to monitor the population size of the insect pest known as the gypsy moth (Lymantria dispar)

Elden [556K]

Answer: Option A: The number of trees sampled.

The accuracy can be understood as how close is the measured value to the true value. The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.

5 0

3 years ago

Read 2 more answers