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strojnjashka [21]
4 years ago
10

Could someone help me with this one please! Will give brainiest (if possible)! I think it's C. but I'm not sure :/

Physics
1 answer:
const2013 [10]4 years ago
5 0

I believe you are right! If the wheels were bigger then they would add more mobility to the wagon with less effort because Allowing it to take more weight (from the wagon) allowing you to pull much easier.

Have a blessed day!

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A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
4 years ago
Why you so sussy bakugou?
ruslelena [56]

Answer:

IM CRYIFNNNAEWJN

Explanation:

6 0
3 years ago
Read 2 more answers
Justin Bieber is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.
sveticcg [70]

===>  Distance fallen from rest in free fall =

                                         (1/2) (acceleration) (time²)

                               (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²):   (122.5 m / 4.9 m/s²)  =  time²

                                                           (122.5/4.9) s²  =  time²

Take the square root of each side:    5.0 seconds


===> (Accelerating at 9.8 m/s², he will be dropping at
                               (9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'.  We'll need this number for the last part.)


===> With no air resistance, the horizontal component of velocity
doesn't change.

Horizontal distance = (10 m/s) x (5.0 s)  =  50 meters .

===>  Impact velocity =  (10 m/s horizontally) + (49 m/s vertically)

                                 = √(10² + 49²)  =  50.01 m/s  arctan(10/49)

                                 =    50.01 m/s   at  11.5° from straight down,
                                                           
away from the base of the cliff.  

7 0
3 years ago
Read 2 more answers
What is the meaningful of physics ​
AleksandrR [38]

Answer:

Physics is a branch of science. It is one of the most fundamental scientific disciplines. The main goal of physics is to explain how things move in space and time and understand how the universe behaves. It studies matter, forces and their effects. The word physics comes from the Greek

3 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
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