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Lera25 [3.4K]
3 years ago
9

A 43 kg box is being pushed and pulled by

Physics
1 answer:
m_a_m_a [10]3 years ago
4 0

Answer:

a = - 0.209 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

We will take the positive forces to the right and the negative forces to the left.

155-277+113=43*a\\-9=43*a\\a = - 0.209[m/s^{2} ]

The negative sign means that the box accelerates in a negative direction to the left.

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Which method of separation would you use when working in the garden and trying to separate the dirt from the weeds?
Thepotemich [5.8K]

Sifting is the best method cuz all the dirt will be carried by wind.

3 0
3 years ago
Read 2 more answers
The Sun is a star that is in a stable state called the main sequence.
nlexa [21]
Astronomers believe some stars are moving away from earth by measuring the extent to which their light is “stretched” into the red part of the color spectrum, representing the lower frequencies on that spectrum.

In short, stars moving away from earth are “red-shift”, while if they were moving towards the earth they would take on a “blue-shift”.
5 0
2 years ago
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
3 years ago
If you are in an elevator that speed up then the apparent weight is
Sholpan [36]

Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

W_{net} = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

Apparent weight (ma) = W_{net} - mg

3 0
2 years ago
when do you use cos and sin in situations like these? is horizontal always cos and vertical always sin?
Andreas93 [3]

Answer:

yes

Explanation:

this is simple

the horizontal line is adjacent

the vertical line is opposite

recall that cos x=adj/hyp

adj=hyp(cos x)

while opp=hyp(sin x)

8 0
3 years ago
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