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2 years ago

What is the velocity at 10 seconds? PLEASE HELP!!!!!

Physics

2 answers:

Mamont248 [21]2 years ago

7 0

15 explanation: look at the graph

Helen [10]2 years ago

3 0

Answer:

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You might be interested in

At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?

Ipatiy [6.2K]

The resistance of the sample is $682.5\Omega$

Explanation:

The relationship between resistance of a material and temperature is given by the equation

$R(T)=R_0(1+\alpha (T-T_0))$

where

$R_0$ is the resistance at the temperature $T_0$

$\alpha$ is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

$R_0 = 525 \Omega$ when the temperature is $T_0 = 20^{\circ}C$

While the temperature coefficient of resistance of nickel is

$\alpha = 0.006/^{\circ}C$

Therefore, the resistance of the sample when its temperature is

$T=70^{\circ}C$

$R=(525)(1+0.006(70-20))=682.5 \Omega$

Learn more about resistance:

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brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

3 0

3 years ago

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

2 years ago

A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?

kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

= 1092 ft³

The weight of the air = density x volume

= 0.07 lbs/ft³ x 1092 ft³

= 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0

2 years ago

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

Wewaii [24]

Given :-

An elevator is moving vertically up with an acceleration a.

To Find :-

The force exerted on the floor by a passenger of mass m .

Solution :-

As the man is in a accelerated frame that is non inertial frame , we would have to think of a pseudo force .

The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

$\implies Weight_{apparent}= mg + ma$

Hence option d is correct choice .

I hope this helps .

3 0

2 years ago

The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.2 × 10-15 m. The single electron in a hydrogen

Nutka1998 [239]

Answer: 0.86 × 10^14

Explanation:

Given the following :

Radius of proton = 1.2 × 10-15 m

Radius of hydrogen atom = 5.3 × 10-11 m

Density of proton could be calculated thus:

Mass of proton = 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45

Density = mass / volume

Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)

= 0.2306 × 10^18

Density of hydrogen atom:

Mass of hydrogen atom= 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31

Density = mass / volume

Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)

= 0.2676 × 10^4

Ratio is thus:

Density of proton / density of hydrogen atom

0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14

6 0

3 years ago