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3 years ago

If a fish experiences an acceleration of 4.9 m/s over a period of 1.84 a , what was the change in velocity

Physics

1 answer:

Veseljchak [2.6K]3 years ago

5 0

Explanation:

they'll dewwel jss q.v hewn red f quill DC noo

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A 10.-kilogram object, starting from rest, slides down a frictionless incline with a constant acceleration of 2.0 m/sec2 for fou

aniked [119]

Answer:

C) 100N

Explanation:

Formula for calculating the Weight of an object is expressed as;

Weight = mass × acceleration due to gravity

Given

Mass of the object = 10kg

Acceleration due to gravity = 9.81m/s²

Substitute into the formula above

Weight = 10×9.81

Weight = 98.1N

Hence the approximate weight of the object is 100N

5 0

3 years ago

Use Coulomb’s law to derive the dimension for the permittivity of free space. <br><br><br><br>

Paraphin [41]

Answer: M^-1 L^-3T^4A^2

Explanation:

From coloumb's law

K = q1q2 / (F × r^2)

Where;

q1, q2 = charges

k = constant (permittivity of free space)

r = distance

Charge (q) = current(A) × time(T) = TA

THEREFORE,

q1q2 = (TA) × (TA) = (TA)^2

Velocity = Distance(L) / time(T) = L/T

Acceleration = change in Velocity(L/T) / time (T)

Therefore, acceleration = LT^-2

Force(F) = Mass(M) × acceleration (LT^-2)

Force(F) = MLT^-2

Distance(r^2) = L^2

From ; K = q1q2 / (F × r^2)

K = (TA)^2 / (MLT^-2) (L^2)

K = T^2A^2M^-1L^-1T^2 L^-2

COLLEXTING LIKE TERMS

T^2+2 A^2 M^-1 L^-1-2

M^-1 L^-3T^4A^2

3 0

3 years ago

Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a

Ira Lisetskai [31]

Answer:

Explanation:

Given

mass of ethanol $m_e=310\ gm$

mass of aluminium cup $m_{al}=90\ gm$

both are at an initial temperature of $T_i=30^{\circ}C$

specific heat of ethanol $c_e=2.46\ J/g-K$

specific heat of aluminium $c_{al}=0.9\ J/g-K$

specific heat of ice $c_i=2.108\ J/g-K$

specific heat of water $c_w=4.184\ J/g-K$

Latent heat of fusion $L=334\ J/gm$

suppose m is the mass of ice added

Heat loss by Al cup and ethanol after $18^{\circ}C$ is reached

$Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)$

Heat gained by ice such that ice is melted and reached a temperature of $18^{\circ}C$

$Q_2=m\times 2.108\times (8.5)+m\times 334$

Comparing 1 and 2 we get

$m=23.65\ gm$

Thus 23.65 gm of ice is added

8 0

3 years ago

All magnetic fields result from the movement of

konstantin123 [22]

Don’t know sorry I’m just trying not a good person

5 0

3 years ago

25 points!

Veronika [31]

Refer to the diagram shown below.

At A, the boy begins walking up the stairs.

At B, the boy is at the top of the slide. He has acquired PE (potential energy).
The value of the PE is
(50 kg)*(9.8 m/s²)*(11.5 m) = 5635 J

At C, the boy has KE (kinetic energy).
The value of the KE is
(1/2)*(50 kg)*(12 m/s)² = 3600 J

Energy is lost between B and C due to friction.
The lost energy is
5635 - 3600 = 2035 J

The distance traveled along the slide is 108 m.
If F = the average frictional force, then
(F N)*(108 m) = 2035 J
F = 18.84 N

Answers:
(a) The mechanical energy lost by sliding is 2035 J.
(b) The average frictional force is 18.84 N

8 0

3 years ago