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svetoff [14.1K]
3 years ago
12

Can someone help with thsi? i will give brainliest

Physics
1 answer:
kogti [31]3 years ago
4 0

Answer:

40 meters. look for the dot above the 20 on the x-axis and follow it over to the left.

Explanation:

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A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c
bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.

Given an arbitrary x and its double, 2x, launch velocities are

v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.

7 0
3 years ago
An airplane flies 40km directly south in 10 minutes and 20 km directly east in five minutes, its average speed is:
blondinia [14]
Average speed = total distance / total time
total distance = 40 + 20 = 60km
total time taken = 10 + 5 = 15 minutes
Average speed = 60/15 = 4km/min
8 0
3 years ago
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3. What are the challenges of looking for Dyson spheres?
S_A_V [24]

1. it is difficult to search for it . Because infrared rays will never penetrate through earth atmosphere.

2. we are unaware of how it looks like and we only know it is red and will glow . A damaged star also looks like this.

3. Dust also makes is hard to detect Dyson spheres . So we will get confused between Dyson sphere and a star surrounded by dust.

5 0
3 years ago
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A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu
True [87]

Answer:

C

Explanation:

they both have to be the same for both to not move

8 0
2 years ago
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: A honeybucket man is carrying his load. He has a pole 2 m long with a bucket hanging from each end. The buckets have a mass of
nasty-shy [4]

Answer:

Explanation:

Using the principle of moment, assuming the rod is uniform rod of mass 1 kg

the center of mass of the rod will be at 1 m

assuming the system is in equilibrium,

clockwise moment = anticlockwise moment

let the distance of the man shoulder be x from the center of gravity and also is the pivot point

total mass of bucket + mass of honey = 2kg + 3 kg =  5 kg for rear bucket and

2kg + 5 kg = 7 kg for front bucket

( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)

5 + 5 x + x = 7 - 7x

5 + 6x = 7 - 7x

6x + 7x = 7 - 5

13x = 2

x = 2 / 13 = 0.154 m

the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).    

8 0
3 years ago
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