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Taya2010 [7]
4 years ago
14

How can you find the net force if two forces act in opposite direction?

Physics
2 answers:
Fed [463]4 years ago
8 0

Answer:

Newton's third law states that when two bodies interact, the forces are equal and opposite in magnitude (distance and direction).

Explanation:

For example, if a box has 3 Newtons for force pushing to the left, and 3 Newtons pushing to the right, the net force would be zero, because they cancel each other out. But, if there is 3 Newtons pushing left and 2 Newtons pushing right, the net force would be 1 Newton to the left.

faust18 [17]4 years ago
5 0
Subtract one from the to find the magnitude. Whether it's positive or negative will determine the direction.
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A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.50 m above. The second student cat
Mkey [24]

Answer:

a) 8.58 m/s upward

b) -2.211 m/s downward

Explanation:

Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.

(a)We have the following equation of motion

s = v_0t + gt^2/2

where v_0 is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m

3.5 = v_01.1 - 9.81*1.1^2/2

3.5 = 1.1v_0 - 5.94

1.1v_0 = 3.5 + 5.94 = 9.44

v_0 = 9.44 / 1.1 = 8.58 m/s

So the keys were thrown initially upward with a speed of 8.58 m/s

(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is

v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s

So they keys would have a downward speed of 2.211 m/s

5 0
3 years ago
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density
blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

6 0
3 years ago
A 8.0-cm long solenoid has 2000 turns of wire and carries a current of 5.0-A. Calculate the strength of the magnetic field at th
CaHeK987 [17]

Answer:

B = 0.157 T

Explanation:

Given that,

Length of the solenoid, l = 8 cm = 0.08 m

Number of turns in the wire, N = 2000

Current, I = 5 A

We need to find the strength of the magnetic field at the center of the solenoid. It is given by the formula as follows :

N=\mu_o nI, N is number of turns per unit length of solenoid.

So,

B=4\pi \times 10^{-7}\times \dfrac{2000}{0.08}\times 5\\\\=0.157\ T

So, the magnetic field at the center of the solenoid is 0.157 T.

6 0
3 years ago
What is a diaganosis?
vladimir2022 [97]

Answer:

c

Explanation:

5 0
3 years ago
A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the e
mash [69]

Answer:

(a) V_{avg}=2.868m/s

(b) V_{avg}=5.352m/s

(c) V_{avg}=7.836m/s

Explanation:

Given data

x(t)=αt²-βt³

α=1.53m/s²

β=0.0480m/s³

First we need to find distance x at these time so

x(t)=1.53t²-0.0480t³

at t=0

x(0)=1.53(0)²-0.0480(0)³=0m

at t=2

x(2)=1.53(2)²-0.0480(2)³=5.736m

at t=4s

x(4)=1.53(4)²-0.0480(4)³=21.408 m

For(a) Average velocity at t=0s to t=2s

The average velocity is given as

Vavg=Δx/Δt

V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\  As\\x(0)=0m\\x(2)=5.736m\\V_{avg}=\frac{5.736m-0m }{2s-0s}\\V_{avg}=2.868m/s

For(b) Average velocity at t=0s to t=4s

The average velocity is given as

Vavg=Δx/Δt

V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\  As\\x(0)=0m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-0m }{4s-0s}\\V_{avg}=5.352m/s

For(c) Average velocity at t=2s to t=4s

The average velocity is given as

Vavg=Δx/Δt

V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\  As\\x(2)=5.736m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-5.736m }{4s-2s}\\V_{avg}=7.836m/s

8 0
3 years ago
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