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liraira [26]
3 years ago
12

25 PTS

Physics
1 answer:
yarga [219]3 years ago
7 0
The answer would be C
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A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj
Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec     horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0
2 years ago
15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m
Rudik [331]

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters



7 0
3 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r
Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

F_{c} = \dfrac{mv^{2}}{r}

where, m is the mass of the object, v is the velocity of the object and r is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0
3 years ago
How do scientists determine the age of the solar system?
Sholpan [36]
By studying meteorites which are the most ancient material in space.
7 0
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