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Helga [31]
3 years ago
10

Một bình có thể tích 0,5m^ chứa không khí ở áp suất dư 2 bar, nhiệt độ

Physics
1 answer:
Tomtit [17]3 years ago
8 0

Answer:

bình có độ chân không 420 mmHg trong điều kiện nhiệt độ khí xem

như không đổi. Áp suất khí quyển là 768 mm Hg ở 18.C, p = 29kg.

Explanation:

इ दोंत उन्दाएर्स्रंद

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A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.
jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m  </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

4 0
4 years ago
Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65
guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, \theta =65°

Range of the projectile is, R=520 m.

Acceleration due to gravity, g=9.8\ m/s^2

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

R=\frac{v_{0}^2\sin2\theta}{g}

Plug in all the given values and solve for minimum speed, v_0.

520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0
4 years ago
A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl
weeeeeb [17]

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

             

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = \frac{k \Delta x}{ m g}

            cos θ = \frac{880 \ 0.1}{ 150 \ 9.8}

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = \frac{mg \ cos \  \theta}{k}

            Δx = \frac{ 150 \ 9.8 \ cos45}{880}

            Δx = 1.18 m

7 0
3 years ago
A flea jumps straight up to a maximum height of 0.400 m . what is its initial velocity v0 as it leaves the ground?
timama [110]

For an object`s motion, the Kinematic equation is,

v^2=v_{0}^2+2ah

Here, v is the final velocity and h is stands for the height of the object and a is the acceleration of the object.

As according to question,

v=0m/s,a=g-9.8 m/s^2 and h = 0.400 m

Thus, putting these values in above equation, we get

0= v_{0}^2 -2gh

or

v_{0} =\sqrt{2 \times 9.8 \times 0.400 }

v_{0} = 2.8 m/s

Therefore, initial velocity is 2.8 m/s



8 0
3 years ago
ou are looking for a new tennis partner. Which of these people would most likely demonstrate good sportsmanship based on the inf
Dennis_Churaev [7]
<h2><u>Answer:</u></h2>

As you are looking for a new tennis partner. People should keep in mind that they should go for the one who most likely demonstrate good sportsmanship

Luis, when you pursue the principles in tennis, you realize when to talk up, you don't blast a racquet or shout, holler.

Whatever it is following the principles and being respectful it the most ideal approach.

7 0
3 years ago
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