Frequency= speed/ wavelength
=0.5m/s divided 0.1.m
=5.0 Hz
The answer would be letter D.
Answer:
769,048.28Joules
Explanation:
A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How much energy was lost to air friction during this bump
The energy lost due to friction is expressed using the formula;
Energy lost = Potential Energy + Kinetic Energy
Energy lost = mgh + 1/2mv²
m is the mass
g is the acceleration due to gravity
h is the height
v is the speed
Substitute the given values into the formula;
Energy lost = 56(9.8)(1400) + 1/2(56)(5.10)²
Energy lost = 768,320 + 728.28
Energy lost = 769,048.28Joules
<em>Hence the amount of energy that was lost to air friction during this jump is 769,048.28Joules</em>
Answer:
The answer is Insulator, Conductor
Explanation:
A/An Insulator is a material in which charges will not move easily, whereas a/an Conductor is a material that allows charges to move about easily
Answer:
The tension in the two ropes are;
T1 = 23.37N T2 = 35.47N
Explanation:
Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity
W = 4.2×10 = 42N
The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.
Using triangular law of force and sine rule to get the tension we have;
If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;
T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}
T1/sin33.8° = T2/sin57.6° = 42/sin88.6°
From the equality;
T1/sin33.8° = 42/sin88.6°
T1 = sin33.8°×42/sin88.6°
T1 = 23.37N
To get T2,
T2/sin57.6°= 42/sin88.6°
T2 = sin57.6°×42/sin88.6°
T2 = 35.47N
Note: Check attachment for diagram.
Answer:
<h2>8.78 Amps</h2>
Explanation:
Given data:
power rating of the heater P= 1010 W
voltage of the heater V= 115 volts
current taken by the heater I= ?
We can apply the power formula to solve for the current in the heater
i.e P= IV
Making I the current subject of formula we have
I= P/V
Substituting our given data into the expression for I we have
I=1010/115= 8.78 A
<h2 /><h2><em>Hence the current when the unit/heater is operating is 8.78 Amp</em></h2>