All categories

4 years ago

You push a 1.00 food tray through the cafeteria One

Physics

1 answer:

lawyer [7]4 years ago

7 0

Answer:

Acceleration of tray and carton = 6.03 m/s²

Horizontal force exerted by tray in carton = 3.02 N

Explanation:

Given:

Mass of tray (M) = 1.00 kg

Mass of milk carton (m) = 0.50 kg

Force applied on the system (F) = 9.04 N

The tray and carton slide horizontally without friction.

The free body diagram of the system is shown below.

The net force acting on the tray and carton system is the constant force 9.04 N.

The total mass of the system is (M + m) = 1.00 kg + 0.50 kg = 1.50 kg

Now, according to Newton's second law:

Net Force = mass × acceleration

⇒ Acceleration = Net force ÷ mass

Therefore, the acceleration of the tray and carton is given as:

$a=\frac{9.04}{1.50}=6.03\ m/s^2$

Now, consider the free body diagram of the carton alone.

The forces acting on the carton is only the horizontal force (f) exerted by the tray.

So, from Newton's second law:

Net Force = mass × acceleration

$f=ma\\f=0.50\times 6.03\\f=3.02\ N$

Therefore, the horizontal force exerted by the tray on the carton is 3.02 N.

You might be interested in

What is the sound intensity level in a car when the sound intensity is 0.525 μW/m2 ? Use I0 = 1.00×10−12 W/m2 for the reference

never [62]

Answer:

The sound intensity level in the car is 57.2 dB.

Explanation:

Sound intensity level in decibels, β = 10 log (I/I₀); where I = 0.525 × 10⁻⁶ W/m², I₀ = 1.0 × 10⁻¹² W/m²

β (dB) = 10 log ((0.525 × 10⁻⁶)/(1.0 × 10⁻¹²)) = 10 × 5.72 = 57.2 dB

Hope this Helps!!!

8 0

3 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

6 0

3 years ago

HEY CAN ANYONE PLS PLS PLS HELP ME OUT IN DIS I AM STRUGGLING TOO MUCH

meriva

<h3>Answer: 104.5 cubic cm</h3>

=======================================================

Work Shown:

r = radius = 1.045 cm

h = height = 30.48 cm

pi = 3.141 approximately

V = volume of cylinder

V = pi*r^2*h

V = 3.141*(1.045)^2*30.48

V = 104.547940002

V = 104.5 cubic cm

6 0

3 years ago

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B 

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0

3 years ago

A circle has a radius of 13m Find the length of the arc intercepted by a central angle of .9 radians. Do not round any intermedi

Oksi-84 [34.3K]

Answer:

11.7 m

Explanation:

The radius of the circle is 13 m.

The central angle of the arc is 0.9 radians

The length of an arc is given as:

L = r θ

where θ = central angle in radians = 0.9

=> L = 0.9 * 13 = 11.7 m

5 0

3 years ago