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JulijaS [17]
4 years ago
6

A horse is grazing for food. It trots rightward 15 m to eat a carrot, then walks rightward 20 m, to another carrot, then turns l

eftwards to walk 4m, to bite an apple. The horse walks a total time of 74 seconds. What is the magnitude of its average velocity?
(** HINT- you may want to draw this out, remember with VELOCITY direction matters-opposite vectors cancel out-with SPEED direction does not matter).
Physics
1 answer:
mafiozo [28]4 years ago
6 0

Answer:

<h3>0.42s</h3>

Explanation:

Velocity = Displacement/time

Displacement of the horse is the distance covered in a specified direction

Total distance of the horse towards the right = 15m + 20m = 35m

Total distance of the horse towards the left = 4m

Displacement = Distance towards the right - Distance towards the left

Displacement = 35m-4m

Displacement =  31m

Time taken = 74seconds

substitute the values gotten into the velocity formula

average velocity = 31m/74s

average velocity = 0.42m/s

Hence the magnitude of its average velocity is 0.42s

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nasty-shy [4]

Most likely climbing up the mountain

6 0
3 years ago
Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively, kept at a distance of 10 m from
NNADVOKAT [17]

Answer:

Answer

F=G×

d

2

m×M

m = 50 kg

M = 120 kg

Distance, d = 10 m

G=6.7×10

−11

Nm

2

/kg

2

F=6.67×10

−11

×

10

2

50×120

F=6.67×60×10

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F=4.02×10

9

N

6 0
3 years ago
Read 2 more answers
A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling
Veronika [31]

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car \Delta m_ {s} , this happens during a time interval

[t, t + \Delta t], in addition to freight car and sand at time t.

In this way we need to define the two states:

State 1,

consider t, m_ {c} (t) + \Delta m_ {s} and V.

State 2,

consider t + \Delta t, m_ {c} (t), V + V \Delta V

In this state is the mass of sand output, which

is composed of

\Delta m_{s}, V + \Delta V

In this way we define the Linear movement in x, like this:

p_ {x} (t) = (\Delta m_ {s} + m_ {c} (t)) v

p_ {x} (t+\Delta t) = (\Delta m_ {s} + m_ {c} (t)) (v + \Delta v)

m_ {c} (t) = m_ {c, 0} - bt = m_ {c} + m_ {s} -bt

In this way we proceed to obtain the Force

F =\lim_{\Delta t \rightarrow 0} \frac {p_x (t + \Delta t) -p_ {x} (t)} {\Delta t}

F = lim_{\Delta t \rightarrow 0} m_ {c} (t) \frac {\Delta v} {\Delta t} + lim_{\Delta t \rightarrow 0} m_ {s} (t) \frac {\Delta v} {\Delta T}

Since the mass of the second term becomes 0, the same term is eliminated, thus,

F = m_ {c} (t) \frac {dv} {dt}

\int\limit ^ {v (t)} _ {v = 0} dv = \int\limit^t_0 \frac{Fdt} {m_ {c} + m_ {s} -bt}

V (t) = - \frac {F} {b} ln (\frac {m_c + m_s-bt} {m_c + m_s})

3 0
3 years ago
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
Fynjy0 [20]

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

\Delta L=\dfrac{PL}{AE}

A=\dfrac{PL}{\Delta LE}

A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

4 0
3 years ago
Read 2 more answers
3. How does the centripetal force change when
OLga [1]

Answer:

The centripetal force increases.

Explanation:

Centripetal force is given by F_c=ma_c; where a_c=\frac{v^2}{r}.

Given these two equations, we can rewrite the formula for centripetal force as:

F_c=m\frac{v^2}{r}

As shown, radius is in the denominator, thus, a smaller radius would lead to a larger centripetal force.

7 0
3 years ago
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