Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force

Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

$T = \frac{2 u Sin\theta }{g}$

$T = \frac{2 \times 20 \times Sin35 }{9.8}$

T = 2.34 second

(b) The formula for the maximum height is given by

$H = \frac{u^{2} \times Sin^{2}\theta }{2g}$

$H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}$

H = 6.71 m

(c) The formula for the range is given by

$R = \frac{u^{2} \times Sin 2\theta }{g}$

$R = \frac{20^{2} \times Sin 2 \times 35}{9.8}$

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0

3 years ago

What is the equivalent resistance between the points A and B of the network?

Dominik [7]

Explanation:

First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.

3 0

3 years ago

John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0

cestrela7 [59]

Answer:

(a) The horizontal ground reaction force $F_{g,x}=325\, N$

(b) The vertical ground reaction force $F_{g,y}=696\, N$

(c) The resultant ground reaction force $F_g=768\, N$

Explanation:

Given

John mass , m = 65 kg

Horizontal acceleration , $a_x= 5.0 \frac{m}{s^{2}}$

Vertical acceleration , $a_y=0.9 \frac{m}{s^{2}}$

(a) Using Newton's 2nd law in horizontal direction

$F_{g,x}=ma_x$

=> $F_{g,x}=65\times 5\, N=325\, N$

Thus the horizontal ground reaction force $F_{g,x}=325\, N$

(b) Using Newton's 2nd law in vertical direction

$F_{g,y}-mg=ma_y$

=> $F_{g,y}=mg+ma_y$

=> $F_{g,y}=65\times (9.81+0.9)\, N=696\, N$

Thus the vertical ground reaction force $F_{g,y}=696\, N$

(c) Resultant ground reaction force is

$F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}$

=> $F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N$

=> $F_g=768\, N$

Thus the resultant ground reaction force $F_g=768\, N$

5 0

3 years ago

Two red and two overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running throu

PolarNik [594]

Answer:

A. The line leading a bracket around the balls, labeled A, represents the nucleus of the atom.

B. The white line, labeled B, is signaling the first main energy level, as per the quantum model of the atom.

Explanation:

You are describing the model of an atom with two protons, two neutrons, the electron density (orbital), and two electrons.

Then, you want to know what the labels A and B represent.

The nucleus of the atom is at the center, where protons and neutrons are placed. This is supported by three models: Rutherford's model, Bohr's model, and the (currently accepted) quantum model.

Hence, the red balls and the two overlapping balls in the center are representing the protons and neutrons.

The line leading a bracket around the balls, labeled A, represents the nucleus of the atom, since it is signaling both protons and neutrons in the nucleus.

The green fuzzy circular cloud depicts the region where the electrons are: it is fuzzy in correspondence to the quantum model, which states the electrons are not in a fixed position but in a region around the atom. That region is named orbitals. As per the model you cannot tell the exact position of the electron, and that is what the fuzzy cloud means.

In the the Bohr's model that white line, which should be circular, where the electrons are, depicts the orbit of the electrons, and the shell, which were idenfified with letters L, M, N, K. The first shell was the L shell.

In the quantum model, that line corresponds to the principal quantum number n, which is the main energy level. Hence, the line leading to a bracket overlapping the white line depicts the main energy level.

The atom is neutral because it contains the same number of electrons aroud the nucleus as protons inside the nucleus.

3 0

3 years ago

Read 2 more answers

Describe the forces involved in Throwing a spear

Alexeev081 [22]

Explanation:

The center of gravity is near the grip and does not change during throw. "Throwing through the tip," a popular term of how to throw a javelin, means throwing through the grip or center of gravity. The center of pressure is the aerodynamic force of drag and lift on the javelin.

8 0

3 years ago