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ikadub [295]
2 years ago
7

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air

resistance, so the brick is in free fall.(a) How tall, in meters, is the building?(b) What is the magnitude of the brick’s velocity just before it reaches the ground?(c) Sketch ay−t,vy−t, and y-t graphs for the motion of the brick.

Physics
1 answer:
kozerog [31]2 years ago
5 0

Answer:

h=17m

v=18.6 m/s

Explanation: The question can be solved by applying kinematic equations of motion

Data

u=0

a=g

t=1.9 secs

firstly to calculate the height

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

to find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is straight line of equation y=9.8 as acceleration is constant:

Velocity graph is given by y=9.8x ( y as velocity and x as time):

Displacement graph is given by y=4.9x^2 ( x as time, y as displacement):

These graphs are only applicable from x=0 to x=1.9 ... ignore the other graph sections

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For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou
Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0
2 years ago
An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?
Sunny_sXe [5.5K]
Answer is 6.84 approx
reason:-
            (2.78^2+6.25^2)^1/2=6.84 approx
4 0
2 years ago
Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio
Pachacha [2.7K]
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356  N

Ratio = 0.356  N/589.18 N
<em>Ratio = 6.04</em>
5 0
1 year ago
Read 2 more answers
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
2 years ago
Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the
Firlakuza [10]

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

3 0
2 years ago
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