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Stella [2.4K]
3 years ago
9

A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict

ion is 0.23 between the box and the surface.
a) Find the normal force on the box
b) Find the acceleration on the box
Physics
2 answers:
Elan Coil [88]3 years ago
8 0

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

Strike441 [17]3 years ago
5 0

Answer:

a. 235.2 N

b. 22 m/s²

Explanation:

a) As there is no displacement in vertical direction, the normal force would be equal to the weight of the box.

N = mg

N = (24 kg) (9.8 m/s²) = 235.2 N

b) Let the acceleration of the box be a such that net force acting on the box is F.

F = ma

F = applied force - force due to friction

Force due to friction = μ N

F = 585 N - (0.23)(235.2)

m a = 530.9

a = 22 m/s²

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An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see
mash [69]

Answer:

(7.90 × 10⁻¹⁵) J

Explanation:

The electric force exerted on the elecrron by rhe electric field is given by

F = qE

where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C

E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C

F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N

From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron

F = ma

m = mass of the electron = (9.11 × 10⁻³¹) kg

a = acceleration of the electron caused by the electric force = ?

(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

v = final velocity of the electron = ?

a = acceleration of the electron = (5.10 × 10¹⁷) m/s²

y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

Hope this Helps!!!

3 0
2 years ago
Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
2 years ago
A uniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (En
Maru [420]

Answer:

aCM = (2/3)*g*Sin θ

Explanation:

Consider a uniform solid disk having mass M,  radius R and rotational inertia I  about its center of mass, rolling without  slipping down an inclined plane.

In order to get the linear acceleration of the object’s center of mass, aCM ,

down the incline,  we analyze this as follows:

The force of gravity (W = Mg) acting straight down  is resolved into components parallel and  perpendicular to the incline.

Since the object rolls without  slipping there is a force of  friction (Ff) acting on the object,  at it’s point of contact with the  incline, in the direction up  the incline.

Newton’s 2nd Law gives then for acceleration down the incline

∑Fx' = m*aCM   ⇒    m*g*Sin θ - Ff = m*aCM

The force of friction also causes a torque around the center of mass

having lever arm R so we can also write

τ = R*Ff = I*α

Solving for the friction,    Ff = I*α / R

This is used in the expression  derived from the 2nd Law:

m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

The objects angular acceleration is related to the linear acceleration  of the edge that contacts the incline by

a = R*α

Since the object rolls without  slipping this has the same  magnitude as aCM so we have  that

α = aCM / R

Using this in

m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM

⇒  aCM = (m*g*Sin θ*R²) / (I + m*R²)

if I = (1/2)*m*R²   (for a uniform solid disk)

we get

aCM = (2/3)*g*Sin θ

6 0
3 years ago
What is the conclusion of coin and feather experiment? ​
valina [46]

Answer:

So the conclusion is that in presence of air net force acting downward reduces for feather and hence falls slower than coin. But in absence of air resistance, net downward force is just equal to force due to gravity which is same for both coin and feather and hence they fall down at the same rate.

5 0
3 years ago
Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum
marta [7]

Answer:

  • v_{top} = 61.96 \frac{mi}{h}

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

v_{top} = 27.7 \frac{m}{s}

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

1609.34 \ m = 1 \ mi

Now, we can divide by 1609.34 meters on both sides:

\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}

The left sides equals 1, so

1 = \frac{1 \ mi}{ 1609.34 \ m}

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 0.0172120 \frac{mi}{s}

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

1 \ h = 3600 \ s

as the seconds are dividing in the velocity, we know divide by 1 hour.

\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}

1 = \frac{3600 \ s}{ 1 \ h}

and know we just multiply our top speed by this conversion factor

v_{top} = 0.0172120 \frac{mi}{s}  \frac{3600 \ s}{ 1 \ h}

v_{top} = 61.96 \frac{mi}{h}

8 0
3 years ago
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