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ioda
3 years ago
11

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ration g is directed 1 Upward and ts value changes with height 2 Upward and its value changes with height 3 There is no accelaration 4 Downward and its value is constant 5 Downward and ts value changes with height
Physics
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for y, and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

v_{x}=v_{0}cos(45\°)=constant

x=(v_{0}cos(45\°))t

For the y-axis

v_{y}=v_{0}sin(45\°)-gt

y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}

Where v_{0}, is the initial velocity.

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Hot rolled steel is generallythan cold drawn steel stronger b) weaker c. more fatigue tolerant d. tougher ar ield streneth 350 M
Burka [1]

Answer:

b) weaker

Explanation:

Hot rolled steel is generally weaker than the cold drawn steel. The yielding stress of cold rolled steel is higher than than hot rolled steel. At the same time the crystal structure remain the same in cold rolled steel. Giving it a better surface finish and closer tolerances. The strength and hardness of a cold roll steel is higher than  the hot rolled steel  

6 0
3 years ago
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
What are the three main parts that make up GAIA? Briefly explain and describe each.
Lorico [155]

Answer:

The Gaia spacecraft is composed of two sections: the Payload Module and the Service Module. The Payload Module is housed inside a protective dome and contains the two telescopes and the three science instruments. They are all mounted on a torus made of a ceramic material (silicon carbide).

Explanation:

Hope this helps, Have a Great Day!!

6 0
3 years ago
Read 2 more answers
Water drips from the nozzle of a shower onto the floor 193 cm below. The drops fall at regular (equal) intervals of time, the fi
Law Incorporation [45]

Answer: 108.81 cm and 48.66 cm

Explanation:

In this, we have to make sure to keep in mind the Gravity effects on the drops. The drops will accelerate when they fall making them travel faster. This means, the velocity is not constant.

What is know:

Height (h) = 193 cm

Gravity (g) =  981 cm/s^{2}

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

x = V_{0} t + \frac{1}{2} at^{2}    (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity  which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

193 cm = \frac{1}{2}gt^{2}

To search for the time:

t =\sqrt{\frac{2*193cm}{981cm/s^{2} } }

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

x = \frac{1}{2} (981cm/s^{2})(0.471s)^{2}

x =108.81 cm

For the third drop:

x = \frac{1}{2} (981cm/s^{2})(0.315s)^{2}

x = 48.66 cm

8 0
3 years ago
If clouds were made of cotton-candy what is rain
dlinn [17]

Answer:

If clouds were made of cotton candy, I think rain would be soda.

Or liquified sugar.

:)

6 0
2 years ago
Read 2 more answers
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