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mamaluj [8]
2 years ago
7

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 78 m horizontally

away from and below the point of release. What is the speed of the ball just before it strikes the ground
Physics
1 answer:
sertanlavr [38]2 years ago
5 0

Answer:

Error

Explanation:

Speed= Distance / Time

Speed = 0.1

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The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which co
svp [43]

Answer:

2 in front of water and 1 in front of oxygen

Explanation:

3 0
3 years ago
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A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
Why are black holes black?
melomori [17]
Light can't reflect off them I think
8 0
3 years ago
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If a force of 2n does work at the rate of -2sqrt2 w on an object moving iwth a speed of 2m/s the ale between the fore and teh ve
GrogVix [38]

Answer:

The value of the angle theta = 135degree, which is option c.

Explanation :

These are the options given in the question

a) 45 degrees

b)120 degrees

c)135 degrees

d)150 degrees

e) it is impossible to deliver a negative power.

The concept of work done on an object is applied here.

work = Fd

F = ma

work = mad

Work is a form of energy transferred by an object after the application of a force that act on the object. A step by step explanation is as attached below.

6 0
3 years ago
If 30ml of a fluid has a mass of 63g what is it’s density
murzikaleks [220]

Answer:

The answer to your question is 2.1 g/ml

Explanation:

Data

volume = 30 ml

mass = 63 g

density = ?

Process

Density is defined as the mass per unit volume. The units of density are g/ml or kg/m³.

Formula

Density = mass / volume

Substitution

Density = 63 / 30

Result

Density = 2.1 g/ml

7 0
3 years ago
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