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3 years ago

13

Which fundamental force is very strong across a short distance?

2 answers:

lesya692 [45]3 years ago

5 0

B. Strong nuclear

Because the strong nuclear force is the strongest at short distances, it dominates over the other forces and the two protons become bound, forming a helium nucleus (typically a neutron is also needed to keep the helium nucleus stable).

Westkost [7]3 years ago

3 0

Strong nuclear because it’s the strongest force that can cover a short distance

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A 615 N student standing on a scale in an elevator notices that the scale reads 645 N. From this information, the student knows

BARSIC [14]

Answer:

The elevator must be moving upward.

Explanation:

During the motion of an elevator, the weight of the person deviates from his or her actual weight. This temporary weight during the motion is referred to as "Apparent Weight". So, when the elevator is moving downward, the apparent weight of the person becomes less than his or her actual weight.

On the other hand, for the upward motion of the elevator, the apparent weight of the person becomes more than the actual weight of that person.

Since the apparent weight (645 N) of the student, in this case, is greater than the actual weight (615 N) of the student.

<u>Therefore, the elevator must be moving upward.</u>

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3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

4 years ago

Cats are groomed often by their owners to remove hair. What logical inference can be made based on this statement?

Tpy6a [65]

Answer:

its keeps the fur straight.....im kinda confused with the question

Explanation:

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3 years ago

If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is

devlian [24]

326.85 degrees Celsius

7 0

4 years ago

What are the two most important processes in the cycling of oxygen in and out of the atmosphere?

butalik [34]

C.
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5 0

3 years ago

Read 2 more answers