2. Connect concepts on the left with a description on the right.

A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik

shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle $=40^{\circ}$

$x =12\pm 0.27$

y=1.05

equation of trajectory of ball is given by

$y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }$

for x=12.27

$1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }$

u=10.69

for x=11.73

$1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }$

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0

2 years ago

During energy transformation, all energy systems: are

Tom [10]

B: Energy lose

i say this because in order to change they lose energy.

3 0

3 years ago

Read 2 more answers

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini

Volgvan

Answer:

The horizontal range will be $2.55\times 10^5m$

Explanation:

We have given initial speed of the shell u = $1.6\times 10^3m/sec$

Angle of projection = 51°

Acceleration due to gravity $g=9.8m/sec^2$

We have to find maximum range

Horizontal range in projectile motion is given by

$R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m$

So the horizontal range will be $2.55\times 10^5m$

6 0

2 years ago

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0

3 years ago