While a constant total force of 17 n is exerted on a cart, the cart's acceleration is 5 m/s2. find the mass of the cart?

A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

Aleks [24]

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

3 0

3 years ago

Please please please please help

aivan3 [116]

Answer:

I am pretty sure its the second one but I could be wrong sorry if I am.

Explanation:

:D

4 0

3 years ago

Read 2 more answers

A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A

Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength $\frac{\lambda}{2}$ .

Hence, we have

$\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}$

$\lambda=32\text{ cm}=0.32 \text{ m}$ .

The speed of a wave is the product of its wavelength and its frequency.

$v=f\lambda$

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

$f=\dfrac{343}{0.32}=1070 \text{ Hz}$

7 0

3 years ago

Which statement correctly describes the gravitational potential energy of the pendulum based on this diagram?

MrRa [10]

I think it's 'C' but I won't know for sure until you let me see the diagram.

8 0

3 years ago

in a certain pinhole camera, the screen is 10 cm from the pinhole. When the pinhole is placed 6 cm away from a tree ,a sharp ima

vagabundo [1.1K]

Answer:

Height of the tree is <u>9.6 cm</u>

Explanation:

We know that, Magnification of an image is written as follows.

$\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)$

Where,

$\begin{array}{l}{\mathrm{H}_{0}=\text { height of the object }} \\ {\mathrm{H}_{\mathrm{i}}=\text { height of the image }} \\ {\mathrm{D}_{0}=\text { distance of the object }} \\ {\mathrm{D}_{\mathrm{i}}=\text { distance of the image }}\end{array}$

As per given question,

$\begin{array}{l}{\mathrm{H}_{1}=\text { height of the image }=\text { height of the image of the tree on screen }=16 \mathrm{cm}} \\ {\mathrm{D}_{0}=\text { distance of the object }=\text { distance of the tree from the pinhole }=6 \mathrm{cm}} \\ {D_{1}=\text { distance of the image }=\text { distance of the image from the pinhole }=10 \mathrm{cm}} \\ {\mathrm{H}_{0}=\text { height of the object }=\text { height of the tree }}\end{array}$

Substitute the values in the above formula,

$\begin{array}{l}{\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)} \\ {\left(\frac{16}{H_{o}}\right)=\left(\frac{10}{6}\right)} \\ {\mathrm{H}_{\mathrm{o}}=\left(\frac{16 \times 6}{10}\right)} \\ {\mathrm{H}_{\mathrm{o}}=9.6 \mathrm{cm}}\end{array}$

Height of the tree is 9.6 cm.

8 0

3 years ago