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elena-s [515]
4 years ago
5

A sailboat travels a distance of 600 m in 40 seconds. What speed is it going?

Physics
1 answer:
Gekata [30.6K]4 years ago
7 0

Answer:

15 miles /seconds

Explanation:

Distance = 600m

Time = 40 seconds

Speed=?

speed =  \frac{distance}{time}  \\ speed =  \frac{600}{40}

Simplify

\frac{600}{40}  =  \frac{60}{4}  \\  = 15

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Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie
Stella [2.4K]

Answer:

Angle of incline is 20.2978°

Explanation:

Given that;

Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

8 0
3 years ago
Please help me!!!!<br> Describe the roles these organisms play in their ecosystem.
creativ13 [48]

Answer:

Explanation:

Deer = primary consumer

Green plants = producer

Fungi = decomposers

3 0
3 years ago
A 36-cm-circumference loop of wire is placed between the poles of an electromagnet. When a field of 0.76 T is switched on in a t
Marrrta [24]

Answer:

R = 25.3ohms

Explanation:

See attachment below.

5 0
3 years ago
Read 2 more answers
Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini
amid [387]

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2}  } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r

=\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2}  }{2R^{3} } ]

∴NOTE: Graph is attached

8 0
4 years ago
A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d
TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

W=mgh.................(1)

where m is the mass, g is acceleration due to gravity taken as 9.8m/s^2 and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

h=132sin12^o\\h=27.44m

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

W=54*9.8*27.44\\W=14523.55J

4 0
3 years ago
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