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Law Incorporation [45]
1 year ago
9

A 755 N box rests on a floor with a coefficient of kinetic friction of 0.600 and a coefficient of static friction of 0.800. What

horizontal force is required to start the box moving
Physics
1 answer:
GrogVix [38]1 year ago
4 0

Answer:

fr = R× co-efficient

= 755 × 0.80  

= 604N

Explanation:

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Find the angle (above the horizontal) at which a projectile achieves its maximum range, if y=y0.
KatRina [158]
The answer is 45 degrees. 
According to the Kinematics of projectile motion, if the purpose is to maximize range, optimum angle of landing is always 45 degrees.If the purpose is to maximize range & projection height is zero, the optimum angle of projection (and landing) is 45 degrees.
5 0
2 years ago
While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6
Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

Person A drives at a constant speed of 55 mph for the entire trip,

Time taken by person A

             t_A=\frac{S}{55}

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

            t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}

Ratio of time of arrival of B to A

                      \frac{t_B}{t_A}=\frac{\frac{11S}{585}}{\frac{S}{55}}=\frac{121}{117}=1.034

B will take 1.034 times the time of A from Boston to Hartford.

8 0
2 years ago
Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
Burka [1]
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
7 0
2 years ago
Read 2 more answers
block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
2 years ago
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If th
DiKsa [7]
<h2><u>Answer:</u></h2>

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:

1. (There would be variables that would be hard to control, leading to less reliable data.)

3. (Meter sticks may lack precision or may be read incorrectly.)

4. (Real glider data may vary since real collisions may involve loss of energy.)

5. (Human error in recording or plotting the data could be a factor.)


6 0
2 years ago
Read 2 more answers
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