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3 years ago

Signal far enough ahead so other drivers in your vicinity can make adjustments to your change in speed and ___________.

Physics

1 answer:

3241004551 [841]3 years ago

8 0

Answer:

DIRECTION

Explanation:

the answer of the blank will be DIRECTION

As the signal is far ahead so driver can make the decision of what he have to do either he can change direction or he can be on his same path.

But for that driver has to change his or her speed.

so, from the statement given the blank will be filled with DIRECTION

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A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a

lions [1.4K]

Answer:

Approximately $4.61\times 10^{3}\; {\rm N}$ upwards (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ .)

Explanation:

External forces on this astronaut:

Weight (gravitational attraction) from the earth (downwards,) and
Normal force from the floor (upwards.)

Let $(\text{normal force})$ denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

$\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ .

Let $m$ denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be $(\text{weight}) = m\, g$ .

Let $a$ denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be $(\text{net force}) = m\, a$ .

Rearrange $\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ to obtain an expression for the magnitude of the normal force on this astronaut:

$\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}$ .

3 0

2 years ago

Consider a car engine running at constantspeed. That is, the crankshaft of the en-gine rotates at constant angular velocity whil

Irina-Kira [14]

Answer:

The value is $|v| = 7.39 \ m/s$

Explanation:

From the question we are told that

The angular speed is $w = 2030 \ rpm = \frac{2030 * 2 * \pi }{ 60} = 212.61 \ rad/s$

The distance between the minimum and maximum external position is $d = 6.95 \ cm = 0.0695 \ m$

Generally the amplitude of the crank shaft is mathematically represented as

$A = \frac{d}{2}$

=> $A = \frac{0.0695}{2}$

=> $A = 0.03475 \ m$

Generally the maximum speed of the piston is mathematically represented as

$|v| = A * w$

=> $|v| = 0.03475 * 212.61$

=> $|v| = 7.39 \ m/s$

7 0

3 years ago

A cyclist rode an average speed of 10 miles per hour for 15 miles how long was the ride

Ainat [17]

I think it's an hour and a half

7 0

3 years ago

Find the pressure exerted by a force 8000 newton's on an area of 25m^2. Give your answer in newton's/n^2

Vanyuwa [196]

The pressure exerted by this force is equal to 320 $N/m^2$ .

<u>Given the following data:</u>

Force = 8000 Newton.

Area = 25 $m^2$ .

To calculate the pressure exerted by this force:

<h3>How to calculate pressure.</h3>

Mathematically, pressure is given by this formula:

$P=\frac{F}{A}$

<u>Where:</u>

P is the pressure.

F is the force.

A is the area.

Substituting the given parameters into the formula, we have:

$P = \frac{8000}{25}$

P = 320 $N/m^2$ .

Read more on pressure here: brainly.com/question/8033367

5 0

2 years ago

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

4 0

3 years ago