Plz help... asap I need it

At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial

Bess [88]

The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)

(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s

The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s

<em>ANSWER: 19.12 s</em>

5 0

3 years ago

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Question 18 according to newton's second law of motion, the force acting on an object varies directly with the object's accelera

leonid [27]

Solve for "x"
X=force
18/6=x/9
cross multiply
162=6x
x=27
Hope this helps

3 0

3 years ago

You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment

Margarita [4]

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

12 m north = 12 j

6.00 m east = 6 i

Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

5 0

3 years ago

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

3 years ago

ILI

Nikolay [14]

The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

$\text {Kinetic energy}=\frac{1}{2} m v^{2}$

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.

The final kinetic energy is

$\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5$ = 7500 J

As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.

Work done = Final Kinetic energy - Initial Kinetic energy

Thus the work utilized for moving the car is

Work done = 7500 J - 0 J = 7500 J

Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

7 0

3 years ago