Ask question

Ask question

All categories

1 year ago

8

A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa

gon handle, which is inclined at 35° to the sidewalk as shown below. What is the magnitude of the fore of friction on the wagon?

1 answer:

Jlenok [28]1 year ago

5 0

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx = 18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

You might be interested in

Identify the row that contains two scalars and one vector quantity: Distance Acceleration Velocity Speed Mass Acceleration Dista

GenaCL600 [577]

Answer:

Speed, mass and acceleration

Explanation:

A scalar quantity is a quantity that has only magnitude but no direction while a vector quantity has both magnitude and direction.

According to the question, the row that has two scalars and one vector is speed, mass and acceleration.

The two scalars in this row are speed and mass while the vector quantity there is the acceleration.

Acceleration has direction since it possess direction. A body accelerating will do so in a particular direction. Speed and mass doesn't possess any direction. Mass only specify the magnitude of the body but no clue as to which direction is the body moving towards.

Speed also only specify the

total distance covered with respect to time but not the direction of the direction.

8 0

1 year ago

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th

Otrada [13]

Answer:

$8616.7468 \ kg/m^3$

Explanation:

Pressure is measured is $p=\rho gh$ here p is pressure $\rho$ is density and h is height

We have given pressure $p=9.891\times 10^4\ Pa$ acceleration due to gravity $g=9.9870\ m/sec^2$ height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

5 0

1 year ago

Read 2 more answers

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

1 year ago

Read 2 more answers

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

1 year ago

The speed of sound in seawater is 1470 m/s. A dolphin sends out a click that reflects off of an

Nitella [24]

Answer: 0.204 s

Explanation:

The speed of sound $V$ is defined as the distance traveled $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=1470 m/s$ is the speed of sound in seawater

$t$ is the time the sound wave travels from the dolphin and then returns after the reflection

$d=2(150 m)$ is twice the distance between the dolphin and the object to which the sound waves are reflected

Finding $t$ :

$t=\frac{d}{V}$

$t=\frac{2(150 m)}{1470 m/s}$

<u>Finally:</u>

$t=0.204 s$

3 0

2 years ago