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1 year ago

A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.

Physics

1 answer:

enot [183]1 year ago

7 0

Answer:

As Per Provided Information

Moving body has 2m/s² acceleration

Time taken by body is 4 second

We are asked to find the 'change in velocity' ( ∆V) by the body.

Formula Used here

$\boxed{\bf{\Delta \: V \: = acceleration \: \times time \:}}$

Substituting the given value

$\sf\longrightarrow\Delta\:V \: = 2 \times 4 \\ \\ \\ \sf\longrightarrow\Delta\:V \: =8m {s}^{ - 1}$

Therefore,

Change in velocity is 8 m/s

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Identify the labeled parts in the figure.

Hitman42 [59]

Answer:

I. a, c, f and h

II. e

III. b, d, g and i

IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

II. Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e

III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

5 0

2 years ago

when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of

Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂ (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂

or, C₂=2X/m. T₂ (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂ /2X /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0

3 years ago

She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

Reika [66]

Answer:

The boats are 934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i = 1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

Step 1 : Finding the value of x(Refer the figure attached)

We can use the tangent ratio to find the x value

$tan(42^{\circ}) = \frac{1353}{x}$

$x = \frac{1353}{tan(42^{\circ}) }$

x = 590.47 feet

Step 2 : Finding the value of z (Refer the figure attached)

$tan(29^{\circ}) = \frac{1353}{z }$

$z = \frac{1353}{tan(29^{\circ})}$

z = 1525.12 feet

Step 3 : Finding the value of y (Refer the figure attached)

y = z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

3 0

3 years ago

A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K).

mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

$P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa$

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

$P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa$

(c) Same steps as in part (a)

$P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa$

ΔP=9.164-9.14

ΔP=0.0245kPa

8 0

3 years ago

Change 60cmHg to pascal

xenn [34]

Answer:

1cmhg= 1333.22pa

60cmhg= 60×1333.22

60cmhg=79993.2pa

3 0

2 years ago

A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.​

A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.