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Vladimir79 [104]
3 years ago
10

An object is thrown vertically upward at 27.1 m/s. The velocity of the object 3.4 seconds later is ____ m/s. Round your answer t

o the nearest tenth. Do not use scientific notation. Take up as positive and down as negative.
Physics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

-6.2 m/s (downward)

Explanation:

The velocity of an object thrown vertically upward is given by:

v= u + at

where:

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this problem,

u = 27.1 m/s

t = 3.4 s

So, the velocity after 3.4 s is

v=27.1 m/s + (-9.8 m/s^2)(3.4 s)=-6.2 m/s

and the negative sign means the velocity points downward.

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The amount of energy in food is measured in
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The amount of energy in food is measured in calories. 
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A driver traveling at 40 m/s slams on his brakes the vehicle completely stops in five seconds how far to the vehicle go while st
lara [203]

Answer: 100 m

Explanation:

Initial speed(u) = 40 m/s

Final speed(v) = 0 m/s

Acceleration = v-u/t

=> 0-40/5

=> 8 m/s^2

Displacement= s

=> 2as = v^2-u^2

=> 2(-8)(s) = 0 - (40)^2

=> -16s = -1600

s = 100 m

We can say the vehicle move 100 meters before stopping.

3 0
3 years ago
Read 2 more answers
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s . The rocket engine, when it is fired, exer
ivanzaharov [21]

Answer:

The rocket has to be launched 8 m from the hoop

Explanation:

Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).

On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration

    F -W = m a

    a = (F-mg) / m

    a = F/m  -g

    a = 7.0/0.500  - 9.8

    a = 4.2 m/s²

We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations

Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)

    y = v_{oy} t + ½ a t²

    y = 0 + ½ a t²

    t = √ 2y/a

    t = √( 2 15 / 4.2)

    t = 2.67 s

This time is also the one that takes in the horizontal movement, let's calculate how far it travels

    x = v₀ₓ t

    x = 3 2.67

    x = 8 m

The rocket has to be launched 8 m from the hoop

8 0
3 years ago
Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are se
Scorpion4ik [409]

Answer:

5 N

Explanation:

F_1 = 20 N

r_1 = 6 cm

r_2 = 12 cm

k = Coulomb constat

q = Charge

\dfrac{F_1}{F_2}=\dfrac{\dfrac{kq_1q_2}{r_1^2}}{\dfrac{kq_2q_2}{r_2^2}}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{12^2}{6^2}\\\Rightarrow \dfrac{F_1}{F_2}=4\\\Rightarrow F_2=\dfrac{F_1}{4}\\\Rightarrow F_2=\dfrac{20}{4}\\\Rightarrow F_2=5\ N

The force is 5 N

7 0
3 years ago
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