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2 years ago

Do only number 3 and thank

Physics

1 answer:

IrinaK [193]2 years ago

3 0

Charge = 0.2 Ah is the correct answer...

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I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2

olga nikolaevna [1]

Answer:

Explanation:

$a = \frac{v_f-v_0}{t}$ which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator. For us:

$a=\frac{2.7-1.6}{1.2}$ so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0

2 years ago

At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta

Evgesh-ka [11]

727.5256266 AWNSERrRrrRr

3 0

2 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago

he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1

meriva

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

5 0

3 years ago

type 2 diabetes runs in familes, but its frequency has been rising in recent years. what factors might explain this increase in

FrozenT [24]

Answer:

poor diet andlack of exersise

Explanation:

8 0

2 years ago