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3 years ago

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A model rocket is fired vertically from rest. It has a constant acceleration of 16.4 m/s2 for the first 1.50 s. Then its fuel is

exhausted, and it is in free fall. The rocket has a mass of 78.2 g; the mass of the fuel is much less than 78.2 g. Ignore air resistance. What was the downward net force on the rocket after its fuel was spent?

1 answer:

il63 [147K]3 years ago

3 0

Answer:

The force downward acting o the rocket is F=m*g=0.0782kg*9.8m/s^2=0.76636 N

Explanation:

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Distance covered by Anthony is 3 km. Don't know about displacement though.

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. Emily pushes a 38.8 kg grocery cart of groceries by exerting a 76.0 N force on the handle inclined at 40.0 degrees below the h

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Answer:

a) $F_{x} = 58.2 N$

$F_{y} = 48.9 N$

b) a = 1.5 m/s²

Explanation:

a) The horizontal and vertical components of Emily's force can be found knowing the angle and the exerted force.

Since the handle is inclined at 40.0° below the horizontal we have:

$F_{x} = |F|*cos(\theta) = 76.0 N*cos(40) = 58.2 N$

$F_{y} = |F|*sin(\theta) = 76.0 N*sin(40) = 48.9 N$

b) The acceleration of the car can be calculated as follows:

$F_{x} = ma$

We used the horizontal component of the force because the cart is moving in that direction.

$a = \frac{F_{x}}{m} = \frac{58.2 N}{38.8 kg} = 1.5 m/s^{2}$

Hence, the acceleration of the car is 1.5 m/s².

I hope it helps you!

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You are on the south bank of the river in your canoe you need to reach the north bank you know that you can row your canoe at 2

Anna71 [15]

Answer:

(θ) = 60°

Explanation:

Given:

Speed of canoe Vc = 2 m/s

Speed of River Vr = 1 m/s

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Vc (Cosθ) = Vr

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In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way i

lara [203]

Answer:

Approximately $1.44\times 10^3 \; \rm N \cdot m^{-1}$ assuming that the spring has zero mass.

Explanation:

Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.

On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force $\mathbf{F}$ when its displacement is $\mathbf{x}$ , then its force constant would be:

$\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}$ .

The goal here is to find the expressions for $F$ and for $x$ . By Hooke's Law, the spring constant would be ratio of these two expressions.

Let $T$ represent the time period of this oscillation. With the chair alone, the period of oscillation is $T = 1.00\; \rm s$ .

For a simple harmonic oscillation, the angular frequency $\omega$ can be found from the period:

$\displaystyle \omega = \frac{2\pi}{T}$ .

Let $A$ stands for the amplitude of this oscillation. In a simple harmonic oscillation, both $\mathbf{F}$ and $\mathbf{x}$ are proportional to $A$ . Keep in mind that the spring constant $k$ is simply the opposite of the ratio between $\mathbf{F}$ and $\mathbf{x}$ . Therefore, the exact value of $A$ shouldn't really affect the value of the spring constant.

In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time $t$ would be:

$\displaystyle \mathbf{x}(t) = A \cos(\omega \cdot t)$ .

The restoring velocity at time $t$ would be:

$\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)$ .

The restoring acceleration at time $t$ would be:

$\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)$ .

Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time $t$ would be:

$\begin{aligned}& \mathbf{F}(t) \\ &= m(\text{chair}) \cdot \mathbf{a}(t) \\&= -m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)\end{aligned}$ .

Apply Hooke's Law to find the spring constant, $k$ :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= -\left(\frac{-m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)}{A\cos(\omega \cdot t)}\right) \\ &= \omega^2 \cdot m(\text{chair}) \end{aligned}$ .

Again, $\omega$ stands for the angular frequency of this oscillation, where

$\displaystyle \omega = \frac{2\pi}{T}$ .

Before proceeding, note how $A$ was eliminated from the ratio (as expected.) Additionally, $t$ is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to $k$ :

$\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}$ .

Side note on the unit of $k$ :

$\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}$ .

6 0

3 years ago