All categories

3 years ago

Desperate for help! please!!!!!!!!

Physics

1 answer:

lys-0071 [83]3 years ago

8 0

$0.4029 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$

$\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}$

You might be interested in

By counting the number of waves that pass a certain point each second, we will know the _______ of the wave.

Travka [436]

The answer is A. Frequency

6 0

3 years ago

Read 2 more answers

A physicist found that a force of 1.32 N was measured between two charged spheres. The distance between the spheres was 95 cm. C

labwork [276]

The electric force (and the gravitational force too) is inversely proportional
to the square of the distance between the objects involved.

In this question, the distance is increased by a factor of (1.25/0.95) .

So the electric force will change by the factor of (0.95/1.25)² .

The new force is

(1.32 N) · (0.95/1.25)² = 0.762... newton (rounded)

5 0

3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

A bedroom bureau with a mass of 45 kg, including drawers and clothing, rests on the floor. (a) If the coefficient of static fric

sergij07 [2.7K]

Answer: the minimal force that you need to apply to move the bureau is F = 198.45N

Explanation:

If you want to move an object, you need to apply a force that is bigger than the force of the statical friction.

The force of statical friction can be written as.

Ff = k*N

where k is the coefficient of static friction, in this case, k = 0.45, and N is the normal force between the object and the surface.

In this case, the normal force is the weight of the bedroom bureau, this is:

N = m*g = 45kg*9.8m/s^2 = 441N

Then the force is:

Fr = 0.45*441N = 198.45N

This means that the minimal force that you need to apply to move the bureau is F = 198.45N

and after this point, the force of friction will work wit the kinetic coefficient of friction, that usually is smaller than the statical one.

4 0

3 years ago

A Boeing 737 airliner has a mass of 20,000 kg and the total area of both wings (top or bottom) is 100 m2. What is the pressure d

kvv77 [185]

Answer:

The correct option is A = 1960 N/m²

Explanation:

Given that,

Mass m= 20,000kg

Area A = 100m²

Pressure different between top and bottom

Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as

∑Fy = Wp + FL = 0

where

Wp = is the weight of the plane, and

FL is the lift pushing up on the plane.

Let solve for FL since the mass of the plane is given:

Wp + FL = 0

FL = -Wp

FL = -mg

FL = -20,000× -9.81

FL = 196,200N

FL should be positive since it is opposing the weight of the plane.

Let Equate FL to the pressure differential multiplied by the area of the wings:

FL = (Pb −Pt)⋅A

where Pb and Pt are the static pressures on bottom and top of the wings, respectively

FL = ∆P • A

∆P = FL/A

∆P = 196,200 / 100

∆P = 1962 N/m²

∆P ≈ 1960 N/m²

The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct

5 0

3 years ago

Read 2 more answers