A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force

Answers

Answer 1

Answer:

Answer:

F = 6[N].

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

$P=m*v\nor\nP=F*t$

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 10 [kg]

v = velocity [m/s]

F = force [N]

t = time = 5 [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

$(m_(1)*v_(1))+F*t=(m_(1)*v_(2))$

where:

m₁ = mass of the object = 10 [kg]

v₁ = velocity of the object before the impulse = 1 [m/s]

v₂ = velocity of the object after the impulse = 4 [m/s]

$(10*1)+F*5=10*4\n10+5*F=40\n5*F=40-10\n5*F=30\nF=6[N]$

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7) Straws work on the principle of the outside atmospheric pressure pushing the fluid (for example water) up the straw after you have lowered the pressure at the top of the straw (in your mouth). Assuming you could create a perfect vacuum in your mouth, what is the longest vertical straw you could drink water from?

Answers

Answer:

The longest straw will be 10.328 meters long.

Explanation:

The water will rise up to a height pressure due to which will balance the atmospheric pressure.

We know

$P_(atm)=101325N/m^(2)$

Pressure due to water column of height 'h'

$P_(water)=1000* 9.81* h$

Equating both the values we get the value of height 'h' as

$h=(101325)/(1000* 9.81)\n\nh=10.328m$

True Or False for each question

Answers

7 true

8 false

9 false

10 false

11 false

12 true

13 true

hope this helps!

A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).

Answers

Answer:

time is 32 s and speed is 304.3 m/s

Explanation:

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

$h = u t +0.5 at^2\n\n- 146 =14 sin 43 t - 4.9 t^2\n\n4.9 t^2 - 9.5 t - 146 =0 \n\nt =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\n\nt=(9.5\pm 54.3)/(9.8)\n\nt = 32.05 s, - 22.4 s$

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is

v' = u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

$v=√(v'^2 + v''^2)\n\nv = √(304.1^2 +10.3^2 )\n\nv = 304.3 m/s$

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=(I)/(\Delta t)=(328.6 kg m/s)/(0.812 s)=404.7 N$

ou are writing a short adventure story for your English class. In your story, two submarines need to arrive at a place in the middle of the Atlantic Ocean at the same time. They start out at the same time from positions equally distant from the rendezvouspoint. They travel at different speeds, but both go in a straight line. The first submarine travels at an average speed of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. In the story’s plot, the second submarine is required to travel at a constant velocity, so the captain needs to determine the magnitude of that velocity. What is that velocity?

Answers

Answer:

The constant speed of second submarine is 31.16 km/hr

Explanation:

Given that

v₁=20 km/hr ,d₁= 500 Km

v₂=40 km/hr ,d₂=500 km

v₃=30 km/hr, d₃=500 km

v₄=50 km/hr ,d₄=500 km

We know that

Displacement = Velocity x Time

d = v t

Total displacement = Average velocity x Total time

$d_1+d_2+d_3+d_4=V_(avg)\left((d_1)/(v_1)+(d_2)/(v_2)+(d_3)/(v_3)+(d_4)/(v_4)\right)$

Now by putting the values

$2000=V_(avg)\left((500)/(20)+(500)/(40)+(500)/(30)+(500)/(50)\right)$

$V_(avg)=31.16\ km/hr$

So the constant speed of second submarine will be the average speed of first submarine because they have to meet at the same time.

The constant speed of second submarine is 31.16 km/hr

Could you please solve it with shiwing the full work1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.

3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.