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3 years ago

A 5 kg object slows from +6 m/s to +2 m/s. What is the impulse?

Physics

1 answer:

Liula [17]3 years ago

5 0

Impulse on the object: -20 kg m/s

Explanation:

The impulse exerted on an object is a vector quantity equal to the product between the force exerted on the object and the time interval during which the force is applied.

The impulse exerted on an object is also equal to its change in momentum. Mathematically,

$I=\Delta p = m(v-u)$

where:

m is the mass of the object

u is the initial velocity

v is the final velocity

For the object in this problem, we have

m = 5 kg

u = +6 m/s

v = +2 m/s

Therefore, its impulse is:

$I=(5)(+2-(+6))=-20 kg m/s$

The negative sign means the direction of the impulse is opposite to the direction of motion of the object.

Learn more about impulse:

brainly.com/question/9484203

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Which substance is uniquely a product of incomplete combustion of a fossil fuel?

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I can think of two of them:

-- carbon monoxide

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3 years ago

What is the wavelength of violet light that has a frequency of 7.5x10^14

Crank

Λ= V/f
but change it to represent the speed of light, c 
λ= c/f 
c = 3.00 x 10^8 m/s 
Plug in your given info and solve for λ(wavelength) 
λ= 3.00 x 10^8 m/s / 7.5 x 10^14 Hz
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2 years ago

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100 Points !!!!!!!! I WILL GIVE BRAINLIEST

kumpel [21]

Answer: 1. 0.25 m/s squared

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3.=20kg. I did this before.

Explanation:

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2 years ago

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A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

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4 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 9.6 cm

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

2 years ago