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4 years ago

What are three reasons that friction is needed?

1 answer:

4 0

It is used to reduce speed, without friction, nothing would roll or move on the ground, and lastly, it allows things to endure during bombardment. Hope this helps!

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At a stop light, a truck traveling at 10.5 m/s passes a car as it starts from rest. The truck travels at constant velocity and t

ratelena [41]

Answer:

t = 7 sec.

Explanation:

As the car and the truck travel the same distance, assuming a constant acceleration, we can describe the movement of the truck and the car with these equations for this same displacement:

x(truck) = v*t (1)

x(car) = $\frac{1}{2}*a*t^{2} (2)$

As the left sides of (1) and (2) are equal each other, the same must be true for the right sides:

v*t = $\frac{1}{2}*a*t^{2}$

Solving for t, replacing v= 10.5 m/s and a= 3 m/s², we have:

$t = \frac{2*v}{a} = \frac{2*10.5 m/s}{3 m/s2} = 7 sec.$

⇒ t = 7 sec.

3 0

3 years ago

The range is the horizontal distance from the cannon when the pumpkin hits the ground. This distance is given by the product of

jarptica [38.1K]

The answer is 45 degrees. I am not doing a field experiment for you that involves a cannon and a day's work, for 5 points.

3 0

3 years ago

¿Qué proporción deberían guardar los platos de una prensa hidráulica para que, aplicando 40N de fuerza en el plato menor, podamo

9966 [12]

Answer:

El área de la placa más grande es aproximadamente 9,81 veces el tamaño del área de la placa más pequeña

Explanation:

Los parámetros dados son;

La fuerza aplicada sobre la placa más pequeña, F = 40 N

El culo del objeto levantado en la placa más grande, m = 40 kg

El peso del objeto, W = 40 kg × 9,81 m / s² = 392,4 N

Tenemos la presión en el plato pequeño = La presión en el plato más grande

Por lo tanto;

F / A₁ = W / A₂

Dónde;

A₁ = El área del plato pequeño

A₂ = El área de la placa más grande

Por lo tanto, obtenemos;

40 N / (A₁) = 392,4 N / (A₂)

A₂ / A₁ = 392,4 N / (40 N) = 9,81

∴ A₂ = A₁ × 9,81

El área de la placa más grande, A₂ = 9,81 (9,81 m / s² ≈ g) multiplicado por el área de la placa más pequeña A₁.

3 0

3 years ago

If a car used 250,000 W of power to complete a race in 12 s, how much work did the car do?

Reptile [31]

Answer:

3,000,000 Joules!

Explanation:

power=work/t

250000=work/12

work=3,000,000 Joules

4 0

3 years ago

Read 2 more answers

A metal cube, 2.00cm on each side, has a density of 6600 kg/m3. find its apparent weight when it is totally submerged in water.

LekaFEV [45]

Answer: 0.439488 N

Explanation:

The apparent weight of the metal is computed as

apparent weight = weight - weight of the displaced fluid

To compute the weight of the metal, we use the following formula:

$\text{weight} = \rho gV$

where

$\rho = \text{density of the metal = 6,600 kg/m}^3$
$g = \text{gravitational acceleration = 9.81 m/s}^2$
$V = \text{volume of the metal}$

Note that the volume is unknown but we can compute this because the metal is a cube with edge = 2 cm = 0.02 m. So, the volume of the metal is given by

$\text{Volume} = \text{edge}^3
\\ = (0.02)^3
\\ \boxed{\text{Volume = 0.000008 m} ^3}$

Thus, the weight of the metal is computed as

$\text{weight} = \rho gV
\\ = (\text{6,600 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.517968 kg $\cdot $ m/s}^2
\\ \boxed{\text{weight of the metal} = \text{0.517968 N}}$

Next, we compute the displaced weight or buoyancy, which has the following formula

$\text{weight of the displaced fluid} = \rho' gV'$

where

$\rho' = \text{density of the fluid (water) = 1,000 kg/m}^3
\\ g = \text{gravitational acceleration = 9.81 m/s}^2
\\ V' = \text{displaced volume}$

Note that the displaced volume is equal to the volume of the submerged metal. Since the metal has a volume of $\text{0.000008 m} ^3$ , the displaced volume is $\text{0.000008 m} ^3$ .

Thus, the weight of the displaced fluid is calculated as

$\text{weight of the displaced fluid} = \rho' gV'
\\ = (\text{1,000 kg/m}^3)(\text{9.81 m/s}^2)(\text{0.000008 m}^3) \\ = \text{0.07848 kg $\cdot $ m/s}^2 \\ \boxed{\text{weight of the displaced fluid} = \text{0.07848 N}}$

Therefore,

apparent weight of the metal
= weight of the metal - weight of the displaced fluid
= 0.517968 N - 0.07848 N
apparent weight of the metal = 0.439488 N

3 0

3 years ago