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2 years ago

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

Physics

1 answer:

Kamila [148]2 years ago

5 0

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

${v}^{2} - {u}^{2} = 2gh$

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

${30}^{2} - {0}^{2} = 2 \times 10 \times h$

$= > 20h = 900$

$= > h = \frac{900}{20} = 45$

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can somebody help me answer this question? A car went from 110 m/s to 80 m/s in 20 seconds. What was the acceleration of the car

gogolik [260]

Answer:

-1.5m/s²

Explanation:

Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.

So for this question you have:

V_i = 110m/s
V_f = 80m/s
t_i = 0s
t_f = 20s

which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²

4 0

2 years ago

A bullet of 10g strikes a sand bag at a speed of 100 m/s and gets embedded after travelling

Tasya [4]

Answer:

Solving for time :

(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known

2. The unknown constant that we want to solve)

s = (1/2)(u+v)t <--- one of the formulas

from linear motion

s (distance) = 0.05m

u (initial velocity) = 100m/s

v (final velocity) = 0 m/s (it stops)

t (time taken for change in velocity) = to be found

0.05 = (1/2)(100+0)t

t = 0.001 seconds

Solving for the resistant force :

Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.

When the bullet stops :

F net = 0

F r = F imp

F r = (mu -mv)/t

F r = (0.01x100-0.01x0)/0.001

F r = 1/0.001

F r = 1000N

4 0

2 years ago

An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

Two resistors, R1 and R2, are

Aleonysh [2.5K]

Answer:

The value of R₂ is equal to 24.75 ohms.

Explanation:

Given that,

Two resistors, R₁ and R₂, are connected in parallel.

The equivalent resistance is 14.5 ohms

We need to find the value of R₂.

When two resistors are connected in parallel. The equivalent resistance is given by :

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{14.5}=\dfrac{1}{35}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_2}=\dfrac{1}{14.5}-\dfrac{1}{35}\\\\\dfrac{1}{R_2}=0.04039\\\\R_2=\dfrac{1}{0.04039}\\\\R_2=24.75\ \Omega$

So, the value of R₂ is equal to 24.75 ohms.

8 0

2 years ago

The two measurements necessary for calculating average speed are

alisha [4.7K]

The correct answer is Option (C) distance and time

Explanation:

Average speed of any object is defined as the total distance that object travels over the time it takes to travel that distance. In other words, average speed is the total distance divided by the elapsed time.

$Average \thinspace Speed = \frac{Total \thinspace Distance}{Elapsed \thinspace Time}$

Therefore, as you can see in the above equation, the two measurements that are essential for the calculation of the average speed are the (total) distance and the (elapsed) time.

Hence, the correct option is C.

5 0

2 years ago