The mass of the lawn mower can be calculated from the expression force is equal to the product of mass and acceleration. We do as follows:
F = ma
70 = m(1.8)
m = 38.89 kg
The normal force is the upward force perpendicular with the object. For this case it is equal with Wcos(50). We calculate as follows:
Fn = Wcos50 = 38.89(9.81)(cos50) = 245.23 N
Answer:
a) Calcule a frequência em RPM
= 0.6 RPM
b) a velocidade escalar do carro em m/s.
= 20m/s
Explanation:
a) Calcule a frequência em RPM
A fórmula para calcular a frequência é: 1/T
onde T= Tempo (seconds)
T = 100s
A frequência = 1/100s
A frequência = 0.01Hz
em RPM
A fórmula para calcular a frequência em RPM =
1 Hz = 60RPM
0.01Hz =
A frequência em RPM = 0.01Hz × 60
= 0.6 RPM
b) a velocidade escalar do carro em m/s.
A fórmula para calcular a velocidade escalar = diâmetro ou distância (m) ÷ tempo (s)
Diâmetro ou Distância = 2.0km
Converter 2.0km para m
1 km = 1000m
2km =
2 km × 1000m
= 2000m
A velocidade escalar = 2000m ÷ 100s
A velocidade escalar = 20m/s
Answer:
a) Frequency in RPM
= 0.6 RPM
b) Scalar Velocity
= 20m/s
Explanation:
a) Frequently in RPM
Formula : 1/T
Where T= Time (seconds)
T = 100s
= 1/100s
= 0.01Hz
Frequency in RPM =
1 Hz = 60RPM
0.01Hz = 0.01Hz × 60
= 0.6 RPM
b) Scalar velocity
The formula = Diameter or Distance ÷ Time
Diameter or Distance = 2.0km
Convert 2.0km to m
1 km = 1000m
2km =
2 km × 1000m
= 2000m
Scalar Velocity = 2000m ÷ 100s
Scalar Velocity = 20m/s
Answer:
L' = 1.231L
Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

L: width of the barrier
C: constant that includes particle energy and barrier height
You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.
To find the new value of the L' you can write down both situation for T and T', as in the following:

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

Next, you divide the equation (3) into (4), and finally, you solve for L':

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L
Answer: 35*10^3 N/m
Explanation: In order to explain this problem we know that the potential energy for spring is given by:
Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.
We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.
Then the difference of energy for both cases is 7 J so:
ΔUp= 1/2*k* (0.02)^2 then
k=2*7/(0.02)^2=35000 N/m
Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.
Explanation: On Edgenuity!!