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AURORKA [14]
2 years ago
12

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to ge

t to the top of its trajectory?
Physics
1 answer:
olganol [36]2 years ago
8 0

The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s

<h3>What is Trajectory ?</h3>

When any object is thrown from horizontal at an angle θ except 90°, then the path followed by it is called trajectory, the object is called projectile and its motion is called projectile motion.

Here, we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v = 0.

Given :

  • Horizontal speed= 30m/s
  • Vertical Speed= 30 m/s

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy = vy0 + ayt

Plug in our givens

0 = 30  - 10t

solve for t

t = 3 seconds

Hence, The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s

Learn more about Projectile motion here ;

brainly.com/question/11049671

#SPJ1

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A 3047.8 kg truck has lost its brakes coming down a mountain. Fortunately, there is a ramp of thick gravel inclined at 9.5 degre
Yuliya22 [10]

Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

<u>Step 3: </u>Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

<u>Step 4:</u>  What work is done by the truck on the gravel?  

Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ

5 0
3 years ago
how much energy would microraptor gui have to expend to fly with a speed of 10 m/sm/s for 1.0 minute?
Ray Of Light [21]

Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

Where:

P = power (W)

T = time (s)

Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

P = E/t

E = P x t

E = 8.1 x 60

E = 486 Joules

Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.

Learn more about Microraptor gui here brainly.com/question/1200755

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3 0
1 year ago
Brainliest for the first person to answer Which is a characteristic of an electromagnetic wave?
KiRa [710]
The first one. The E and B chatacteristic are perpendicular to eachother. The direction of the wave can be found by the right hand rule.
5 0
3 years ago
A 52.0-kg sandbag falls off a rooftop that is 22.0 m above the ground. The collision between the sandbag and the ground lasts fo
sergiy2304 [10]

Answer:

F = 7,916,955.0N

Explanation:

According to newtons second law

Force = mass * acceleration

Given

mass = 52.0kg

distance S = 22.0m

time t = 17.0 ms = 0.017s

We need to get the acceleration first using the formula;

S = ut+ 1/2at²

22 = 0 + 1/2 a(0.017²)

22 = 0.0001445a

a = 22/0.0001445

a = 152,249.13m/s²

The magnitude of the average force exerted will be;

F = ma

F = 52 * 152,249.13

F = 7,916,955.0N

4 0
3 years ago
What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma
Sindrei [870]

Answer:

E = 1.50 × 10^{8} V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

\mu _b = \frac{B}{2\mu _o}   ...............1

and

energy density due to electric filed is

\mu _e = \frac{\epsilon _o E^2}{2}     ...............2

and here \mu _b = \mu _ e

so that

E = \frac{B}{\sqrt{\mu _o \times \epsilon _o}}      ...................3

put here value and we get

E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}  

E = 3 × 10^{8}  × 0.50

E = 1.50 × 10^{8} V/m

6 0
2 years ago
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