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3 years ago

A shopper pushes a 5.32 kg grocery cart

Physics

2 answers:

Juli2301 [7.4K]3 years ago

7 0

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

$\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}$

$\text { Acceleration }=\frac{58.232}{5.32}$

$\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

maxonik [38]3 years ago

7 0

Answer: 0.53

Explanation: Try looking it up on yahoo.

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What is the wavelength of the radio waves from an FM station operating at a frequency of 99.5 MHz

ser-zykov [4K]

Answer:

Wavelength, $\lambda=3.01\ m$

Explanation:

It is given that,

Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz

We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :

$c=f\lambda$

$\lambda=\dfrac{c}{f}{$

c = speed of light

$\lambda=\dfrac{3\times 10^8\ m/s}{99.5\times 10^6\ Hz}$

$\lambda=3.01\ m$

So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.

3 0

3 years ago

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$

Comparing we will get

$x_A=-x_B\\\\y_B=6x_A$

Substituting in $x_{B}^{2}+y_{B}^{2}=30.25$

$\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904$

So we have

$\vec{A}=0.904\hat{i}$

Magnitude of vector A = 0.904

8 0

3 years ago

The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are l

marshall27 [118]

Answer:

C)the right cord pulls on the pulley with greater force than the left cord

Explanation:

As we can see the figure that B is connected to the right string while A is connected to the left string

Now force equation for B as it will move downwards is given as

$(2m)g - T_B = (2m)a$

similarly for block A which will move upwards we can write the equation as

$T_A - mg = ma$

now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks

so here pulley will rotate clockwise direction

So tension in the right string must be more than the left string

So correct answer will be

C)the right cord pulls on the pulley with greater force than the left cord