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Which question cannot be answered through making measurements?

Bond [772]

A. is the answer for this question

5 0

4 years ago

A 5kg bag falls a verticle height of 10m before hitting the ground.

g100num [7]

Answer:

$u = 7m {s}^{ - 1}$

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

$me(1) = me(2)$

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

$m \times g \times h = \frac{1}{2} \times m \times u {}^{2} \\$

we divide both sides of the equation with mass to make the math easier.

$9.8 \times 10 = \frac{1}{2} \times u {}^{2} \\ \frac{98}{2} = u {}^{2} \\ u { }^{2} = 49 \\ u = 7$

7 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

The breaking car had 10,000 J of kinetic energy before breaking after breaking it had 2000 J of kinetic energy. How much thermal

WINSTONCH [101]

Answer:

8000J

Explanation:

The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.

Kinetic energy loss by car = thermal energy gained by brakes.

∆K.E = ∆T.E ....1

The Kinetic energy loss by car can be expressed as;

∆K.E = K.E1 - K.E2

Initial K.E = K.E1 = 10000J

Final K.E = K.E2 = 2000J

∆K.E= 10000J - 2000J = 8000J

From equation 1,

∆K.E = ∆T.E

∆T.E = 8,000J

thermal energy gain by brakes = 8,000J

8 0

3 years ago

a 5-kg fish swimming at 1 m/s swallows an absent minded 1-kg fish at rest. What is the speed of the large fish immediately afyer

storchak [24]

Answer:

In first case speed of fish will be $\frac{5}{6}$ m/s.

In the second case the speed of fish $\frac{1}{6}$ m/s

Explanation:

Given data :-

Mass of bigger fish ( m₁ ) = 5 kg.

Mass of small fish ( m₂ ) = 1 kg.

Speed of large fish ( v₁ ) = 1 m/s

Mass of bigger fish after eating smaller one = 5 + 1 = 6 kg.

Case - 1

Momentum of bigger fish before eating the smaller fish = m₁* v₁ = 5 * 1 = 5 kg.m/s

Momentum of bigger fish after eating the larger fish = ( m₁ + m₂)*v

v = speed of bigger fish immediately after lunch.

Using the conservation of momentum.

m₁* v₁ = ( m₁ + m₂)*v

5 = 6 * v

v = $\frac{5}{6}$ m/s.

Case -2

Speed of small fish = 4 m/s

Momentum of bigger fish before lunch = 5 kg.m/s

Momentum of smaller fish before lunch = 4*1 = 4 kg.m/s

Net momentum before lunch = 5 - 4 = 1 kg.m/s

Momentum of bigger fish after eating the larger fish = 6 * V

Using the conservation of momentum.

1 = 6 * V

V = $\frac{1}{6}$ m/s.

3 0

3 years ago