d1=_____m

Part B:

d2=______m

Answer:

**Answer:**

**Explanation:**

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

If you travel 2 km north, then travel 5 km south, what is your displacement?

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking controlled braking threshold braking coasting

How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking controlled braking threshold braking coasting

How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

The problem discusses the change in acceleration when a passenger is added to a car. It requires understanding of Newton's second law of motion, force equals mass times acceleration, and then recalculating the acceleration with the passenger added to the total mass.

This problem pertains to Newton's second law of motion, stating that the force applied on an object equals its mass times its acceleration (**F = ma**). Given that the initial acceleration of the Lamborghini Huracan with a driver is 0.80g or 0.80*9.80 m/s², we can calculate the force applied by the car. By multiplying the car's mass (1510 kg) with its acceleration, we will find the force.

Οnce we have the force, we can calculate the new acceleration if the 80 kg passenger rode along. Given that the force is constant, we determine the car's new acceleration by dividing this force with the new total mass (car mass + passenger's mass). So the question ultimately requires an application of the concepts of **force, mass, and acceleration**.

#SPJ1

The new acceleration of the Lamborghini Huracan with an added passenger can be calculated by finding the initial force using the car's **mass **and** acceleration,** and then using this force with the increased mass (original mass + passenger's mass) to find the new acceleration. The new acceleration will be less than the initial acceleration due to the increased mass.

To determine the new acceleration of the **Lamborghini Huracan** with an added passenger, we first calculate the initial force acting on the car. This can be done by using **Newton's second law** which states that Force = mass * acceleration. Initially, the acceleration is 0.80g (where g is acceleration due to gravity = 9.81 m/s²), and the mass is 1510 kg (including the driver). Therefore, the initial force = 1510 kg * 0.8 * 9.81 m/s².

However, when an 80-kg passenger rides along, the total mass becomes 1510 kg + 80 kg = 1590 kg. To find the new acceleration, we keep the** force** constant (as it is not affected by the introduction of the passenger) and rearrange the formula F = m*a as a = F/m. Use the increased mass to find the new acceleration. Please note that the new acceleration will be less than the initial acceleration due to increased mass.

#SPJ2

**Answer:**

**π*R²*E**

**Explanation:**

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒**Flux = E*π*R²**

в. BaN

с. Ва2N3

D. Ba2N

The formula for barium nitride is Ba(NO3)2

**Answer:**

**3.06 seconds time passes before the watermelon has the same velocity **

**watermelon going at speed 59.9 m/s**

**watermelon traveling when it hits the ground at speed is 79.19 m/s**

**Explanation:**

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

**so 3.06 seconds time passes before the watermelon has the same velocity **

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t² .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

**so watermelon going at speed 59.9 m/s**

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

**so watermelon traveling when it hits the ground at speed is 79.19 m/s**

**Answer:**

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

**Explanation:**

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

The motion of **sand **is due to the movement of **conveyor belt**. The **horizontal **distance between the end of the **conveyor **belt and the middle of the **collecting **drum is 2.044 meters.

The equation of **motion **is the relation between the **distance**, velocity, acceleration and time of a moving body.

The **second **equation of the motion for **distance **can be given as,

Here, is the **initial **body, is the **acceleration **of the body due to gravity and is the time taken by it.

Given information-

The **conveyor **is tilted at an angle of **18**° above the horizontal.

The Sand is moved without slipping at the rate of **2** m/s.

The sand is collected in a big drum **5** m below the end of the conveyor belt.

The **horizontal **component of the velocity is given as,

The **vertical **component of the velocity is given as,

Put the value in the above **equation **as,

The horizontal distance between the end of the **conveyor belt **and the middle of the **collecting drum **is,

Thus, the **horizontal **distance between the end of the **conveyor **belt and the middle of the **collecting **drum is 2.044 meters.

Learn more about the **equation **of **motion **here;

**Answer:**

x = 2.044 m

**Explanation:**

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

solving for t

t = 1.075 sec

for horizontal motion

x = 2cos18*1.075

x = 2.044 m