ME Project- Decomposer

For general projectile motion, which of the following best describes the horizontal component of a projectile's acceleration? As

malfutka [58]

Answer:

E ) The horizontal component of a projectile acceleration is zero.

Explanation:

In case of a projectile , force of gravity acts in vertically downward direction so acceleration will act in vertically downward direction . Its direction never changes during course of its journey. So horizontal component of acceleration will always be zero at all points of its journey.

7 0

3 years ago

A scientist hypothesizes that the temperature at which an turtle's egg is incubated

just olya [345]

Answer:

thw temperature of the male will be higher than that of the female.

6 0

3 years ago

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 55 degree above the horizontal. Ball 2 h

juin [17]

Answer:

d_2 = 4d_1

Explanation:

The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by

R = U²sin2θ/g

Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be

d_1 = U²sin2θ/g

Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be

d_2 = V²sin2θ/g

= (2U)²sin2θ/g

= 4U²sin2θ/g

= 4d_1 (since d_1 = U²sin2θ/g)

So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.

4 0

3 years ago

A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator

bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

$\eta = 1 - \frac{T_2}{T_1}$

Where,

$T_2 =$ Temperature of Cold Body

$T_1 =$ Temperature of Hot Body

$\eta$ = Efficiency

According to the statement our values are:

$\eta = 0.3$

$T_2 = 210K$

Replacing we have that

$0.3 = 1- \frac{210}{T_1}$

$\frac{210}{T_1} = 0.7$

$T_1 = \frac{210}{0.7}$

$T_1 = 300K$

Therefore the temperature of the heat source is 300K

6 0

3 years ago

The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters

skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

5 0

3 years ago