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4 years ago

the public is not yet able to purchase hydrogen fuel cell powered cars because engineers have to determine

Physics

1 answer:

grigory [225]4 years ago

3 0

Answer: This would more a practical feedback than scientific.

Explanation: Despite of the protocols and processes scientists need to develop in order to stablish a new source of energy or daily life option, logistic issues are one of the most important things on the "to do list".

Would it be very expensive to acquire this kind of fuel? Is it efficient? Is it dangerous to human life or the environment? What about the negative reactions this brand new fuel could have in society?.

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3 years ago

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(i) 10 m (ii) 20 m (iii) 40 m (iv) 80 m

IRINA_888 [86]

Answer:

20m

420=80m

100

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3 years ago

A cyclist and his bicycle have a combined mass of 88 kg and a combined

SpyIntel [72]

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3 years ago

A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe

Liula [17]

Answer:

a) v₁fin = 3.7059 m/s (→)

b) v₂fin = 1.0588 m/s (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s ⇒ Kin = 0 J

v₁fin = ?

hin = L = 0.7 m

hfin = 0 m ⇒ Ufin = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*hin = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*hin) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s (→)

b) Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒ v₂fin = 1.0588 m/s (→)

7 0

4 years ago

Desperate for help! please!!!!!!!!

lys-0071 [83]

$0.4029 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the box.

Explanation:

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$

$\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}$

8 0

3 years ago