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emmainna [20.7K]

3 years ago

13

1) A plane's velocity increases from 40 m/s to 100 m/s over a 10 second interval. What is the plane's average acceleration for t

his interval? *

1 answer:

Yakvenalex [24]3 years ago

8 0

Answer:

average acceleration = 6 $\frac{m}{s^2}$

Explanation:

Recall that the average acceleration $(a)$ is defined by the change in velocity from an initial velocity $(v_i)$ , to a final velocity $(v_f)$ over the time (t) it took that change to happen. Then, in mathematical terms this is:

$a=\frac{v_f-v_i}{t}$

with our information this becomes:

$a=\frac{v_f-v_i}{t} = \frac{100-40}{10}=6\,\frac{m}{s^2}$

You might be interested in

The speed of light in a material is 0.50

vivado [14]

Answer:

Explanation:

This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum

Let's find the refractive index of the material

n = c / v

n = 3 10⁸ / 0.5 10⁸

n = 6

When the material passes from one medium to another, it must comply with the law of refraction

n₁ sin θ = n₂ sin θ₂

for the angle criticize the angles tea2 = 90

tea = sin⁻¹n₂/ n₁

The vacuum replacement index is n₂ = 1

tea = sin⁻¹ (1/6)

tea = 9.59º

4 0

2 years ago

nolan ryan has the record for having the speediest fastball in baseball. he could pitch one at 148 i/sec. what is that speed in

Klio2033 [76]

The speed of the ball is 101miles/hr.

A mile is a unit of length that is exactly 1,609.344 metres long. Similarly, 5,280 feet or 1,760 yards make up one mile. The mile is an imperial and common US measurement of distance.

We just have to deal with unit conversions.

One mile is 5280 feet, or 1 ft = 0.000189

The speed of the ball in miles per hour is

$\frac{148ft}{1sec} . \frac{1mile}{5280ft} .\frac{60s}{1min} .\frac{60min}{1hr}$

So, the speed of the ball in miles per hour is 101miles/hr.

Learn more about miles here;

brainly.com/question/23245414

#SPJ4

5 0

1 year ago

If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed. A) saturated

kenny6666 [7]

Answer: D)supersaturated

Explanation: Solubility is defined as the amount of solute in grams which can dissolve in 100 g of the liquid to form a saturated solution at that particular temperature.

At $30^0C$ , the solubility of $KI$ is 153g/100 ml.

Thus if 180 grams is dissolved, it contains more amount of solute than it can hold at that that temperature, and thus is supersaturated solution.

A saturated solution is a solution containing the maximum concentration of a solute dissolved in the solvent. The additional solute does not dissolve in a saturated solution.

An unsaturated solution is solution in which the solute concentration is lower than its equilibrium solubility.

A supersaturated solution is one that has more solute than it can hold at a certain temperature.

6 0

3 years ago

Read 2 more answers

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

3 years ago

To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

sp2606 [1]

Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

$W_f = 40 \times 4$

$W_f = 160 J$

Initial compression in the spring is given as

$F = kx$

$4400 = 1100 x$

$x = 4 m$

now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

$160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)$

$160 = 8800 - 1470 - 30 v^2$

$v = 15.45 m/s$

3 0

3 years ago