Ask question

Ask question

All categories

3 years ago

12

If the lily pads are spaced 2.4 m apart and Ferdinand jumps with a speed of 5 m/s taking 0.6 s to go from lily pad to lily pad,

at what angle must Ferdinand make each of his jumps?

1 answer:

Zina [86]3 years ago

8 0

Answer:

$36.87^{\circ}$

Explanation:

v = Velocity of Ferdinand = 5 m/s

$\theta$ = Angle of jump

T = Time taken = 0.6 s

R = Distance between lily pads = 2.4 m

Horizontal range is given by

$R=v_xT\\\Rightarrow R=vcos\theta T\\\Rightarrow cos\theta=\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{2.4}{5\times 0.6}\\\Rightarrow \theta=36.87^{\circ}$

The angle at which Ferdinand make each of his jumps is $36.87^{\circ}$

You might be interested in

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

RSB [31]

Answer:

The coefficient of kinetic is

$u_{k}=0.59$

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

$F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N$

$F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}$

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

$v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }$

So replacing the acceleration can fin the coefficient:

$u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59$

5 0

3 years ago

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

Why do you not come to thermal equilibrium on a cold day

Tju [1.3M]

Two physical systems are in thermal equilibrium if no heat flows between them when they are connected by a path permeable to heat. Thermal equilibrium obeys the zeroth law of thermodynamics. A system is said to be in thermal equilibrium with itself if the temperature within the system is spatially and temporally uniform.

Systems in thermodynamic equilibrium are always in thermal equilibrium, but the converse is not always true. If the connection between the systems allows transfer of energy as heat but does not allow transfer of matter or transfer of energy as work, the two systems may reach thermal equilibrium without reaching thermodynamic equilibrium.

3 0

3 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago

To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo

Setler [38]

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

$P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\ P = -2mv_{i}$

For Beanbag

value of impulse will large if velocity is zero.

$v_{f}=0\\ P = -mv_{i}$

Explanation:

8 0

2 years ago