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Hoochie [10]
3 years ago
10

Can you help me !!!!!!!!!!!!!

Physics
2 answers:
xxTIMURxx [149]3 years ago
7 0
Nothing is there it’s blank
Anna11 [10]3 years ago
3 0
It’s literally a white screen lol
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Please help i need this by today
makkiz [27]

Answer:

the answer is D

Explanation:

if you read the paragraph and question number 4 it tells you the answer wich is d

3 0
3 years ago
What would changing the frequency of a wave do to the wave?
nata0808 [166]
The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.
3 0
3 years ago
What is the resultant of the vectors shown?
ad-work [718]

Answer:

Option B

Explanation:

Option A is the wrong answer because the horizontal vector is in the opposite direction.

Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]

Option D is the wrong answer as the horizontal vector is in the opposite direction.

5 0
2 years ago
A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig
olya-2409 [2.1K]

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs

the horizontal distance covered at this speed is

y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m

4 0
3 years ago
Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?
Sholpan [36]
You will have to redefine the machine's boundaries.
6 0
4 years ago
Read 2 more answers
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