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4 years ago

9

A ball is kicked at an angle of 35 degrees with the ground. What should be the initial velocity of the ball so that it hits a ta

rget that is 30 meters away at a height of 1.8 meters?

1 answer:

pashok25 [27]4 years ago

6 0

Answer:

18.5 M/S

Explanation:

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When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as

Dafna11 [192]

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>

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3 years ago

Derive an expression for the acceleration of the car. Express your answer in terms of D and vt Determine the time at which the s

Bezzdna [24]

Answer:

V(car) = V(truck) at t = Dt/2

acceleration = v(car) = D/t^2

Explanation:

acceleration = v(car) = D/t^2

Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.

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4 years ago

Melvin is traveling south on I-95 at 29 m/s (65 mph) when a deer jumps into his path, 50 m ahead. a. If his reaction time is 0.1

aleksandr82 [10.1K]

Answer:

a. 5.22 meters

b. 2.9 seconds

c. No, Melvin does not hit the deer

Explanation:

The parameters with which Melvin is travelling are as follows;

The speed of Melvin's motion, u = 29 m/s

The distance from Melvin at which the deer jumps into the path = 50 m

a. Distance, d = Velocity, u × Time, t

The time it takes Melvin to react = 0.18 seconds

The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"

∴ d₁ = 29 m/s × 0.18 s = 5.22 m

The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m

b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²

Therefore, we have;

The time it takes "t₂" it takes for him to come to a complete stop given as follows;

y = u + a × t₂

Where;

v = The final velocity after Melvin comes to a complete stop = 0 m/s

By substituting the known values, we have;

0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂

∴ 29 m/s = 10 m/s² × t₂

t₂ = (29 m/s)/(10 m/s²) = 2.9 s

The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s

c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;

v² = u² + 2·a·d₂

Therefore, we have;

0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂

∴ (29 m/s)² = 2 × 10 m/s × d₂

d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m

The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m

Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.

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3 years ago

Acceleration is how quickly an object’s velocity changes.<br><br> true or false

Fynjy0 [20]

This is true. Hope this helped:)

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3 years ago

Read 2 more answers

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

4 years ago