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2 years ago

6. A stomp rocket takes 1.5 seconds to reach its maximum height

Physics

1 answer:

anyanavicka [17]2 years ago

4 0

Answer:

a) 15 m/s

b) 11.25 m

Explanation:

a) when the rocket is approaching to the maximum height , its velocity =0 (assume g= 10 m/s² downwards)

applying motion equations upwards

V = U +at

0 = U + (-10)*1.5

U = 15 m/s (initial velocity)

b) let maximum height be S and V =0 (velocity at highest point)

V² = U² + 2aS

0 = 15² + 2 * (-10)*S

S = 11.25 m

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Type the correct answer in the box. use numerals instead of words. anne has a sample of a substance. its volume is 20 cm3, and i

CaHeK987 [17]

The density of sample is 5 g/cm3

Given:

volume of sample = 20 cm3

mass of sample = 100 grams

To Find:

density of sample

Solution: Density is the measure of how much “stuff” is in a given amount of space. For example, a block of the heavier element lead (Pb) will be denser than the softer, lighter element gold (Au). A block of Styrofoam is less dense than a brick. It is defined as mass per unit volume

density = mass/volume

d = 100/20

d = 5 g/cm3

So, density of sample is 5 g/cm3

Learn more about Density here:

brainly.com/question/1354972

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6 0

1 year ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

2. Which statement is correct about the Krebs cycle?

IRINA_888 [86]

The correct option is

a. Acetyl-CoA combines with a pyruvic acid to make glucose in the Krebs cycle.

Explanation:

The Krebs citric acid cycle happens within the mitochondrial matrix and generates a pool of energy (ATP, NADH, and FADH2) from the oxidization of pyruvate, the tip product of metabolism. Pyruvate is transported into the mitochondria and loses dioxide to make acetyl-CoA, a 2-carbon molecule.

5 0

3 years ago

Read 2 more answers

A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10

irina [24]

First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.

5 0

2 years ago

Read 2 more answers

A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v

Pavel [41]

Answer:

Explanation:

The resonant frequencies for a fixed string is given by the formula nv/(2L).

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e fn+1 – fn = 390 -325

= 65

3 0

3 years ago