Ask question

Ask question

All categories

3 years ago

10

A ball is projected vertically downward at a speed of 4.00 m/s. How far does the bal travel in 1.80 s? What is the velocity of t

he ball at that time?

1 answer:

MAVERICK [17]3 years ago

7 0

Answer:

1) Distance traveled equals 23.1 meters.

2) Final velocity equals 21.658 m/s.

Explanation:

The problem can be solved using second equation of kinematics as

$s=ut+\frac{1}{2}at^{2}$

where

s is the distance covered

u is the initial speed of the ball

a is the acceleration the ball is under

t is time of travel

Applying the given values in the above equation we get

$s=4.0\times 1.8+\frac{1}{2}\times 9.81\times 1.8^{2}\\\\s=23.1meters$

Part 2)

The velocity after 't' time can be obtained using first equation of kinematics.

$v=u+at$

Applying the given values we get

$v=4+9.81\times 1.80\\\\\therefore v=21.658m/s$

You might be interested in

What would happen to the function or purpose of flower if the stigma wasn't sticky?

sashaice [31]

Answer:

The plant would not reproduce because the flower uses the stigma to catch the pollen

8 0

3 years ago

Read 2 more answers

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

6 0

3 years ago

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart

natita [175]

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,

Force of gravity = magnetic force

$mg=ilB$ ..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$ , v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0

3 years ago

PLEASE HELPPPP!!!!

Tresset [83]

you will hear a higher pitch due to a higher frequency.

3 0

3 years ago

Read 2 more answers

Two blocks of masses m1 and m2 are connected to each other on an Atwood machine (the blocks are connected by a string going over

ddd [48]

Answer:211.11 gm

Explanation:

Given

mass of lighter block $m_1=100 gm$

mass of heavier block is $m_2$

acceleration of system $a=3.5 m/s^2$

From diagram

for lighter block

$T-m_1g=m_1a$

$T=m_1(g+a)$

For heavier Block

$m_2g-T=m_2a$

$T=m_2(g-a)$

Equating Tension T

$m_1(g+a)=m_2(g-a)$

$m_2=m_1\cdot \frac{g+a}{g-a}$

$m_2=m_1\cdot \frac{9.8+3.5}{9.8-3.5}$

$m_2=0.1\cdot 2.11$

$m_2=0.2111 kg$

$m_2=211.11 gm$

8 0

3 years ago