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Colt1911 [192]
3 years ago
9

A 5.0 kg object moving at 10 m/s on a frictionless surfaces collides with but does not stick to a 2.0 kg object that is initiall

y at rest. The force on the 2 kg ball has a magnitude of 2000 N and it lasts for 0.01 s. (a) What is the impulse on the 2 kg object
Physics
1 answer:
lilavasa [31]3 years ago
3 0

Answer:

I= 20 i {N.s}

Explanation:

In order to obtain the impulse on the 2 kg ball, you have to apply the equation of Impulse:

I=FΔt

Where I is the impulse vector, F is the net force and Δt is the interval of time when the force is applied.

In this case:

Δt=0.01 s

F= 2000 i N

where i is the unit vector in the x direction.

Replacing the values in the formula:

I=(2000)(0.01)i

Therefore:

I= 20 i {N.s}

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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
3 years ago
3. A penguin waddles 8 m uphill before sliding back down to its friends in 2 seconds. If the penguin ends where it started, what
german

The velocity of penguin as he ends where he started was 0 m/s.

<h3>What is displacement?</h3>

Displacement is the length of straight line joining the initial and final position of the body.

Given is a penguin who waddled 8 m uphill before sliding back down to its friends in 2 seconds.

We know that the velocity is the rate of change of displacement with respect to time. Mathematically -

v = dx/dt

dx = v dt

∫dx = ∫v dt

Δx = vΔt

v = Δx/Δt

Now, the displacement of the penguin will be = Δx = 8 - 8 = 0

Then, its velocity will be -

v = 0/Δt = 0

Therefore, the velocity of penguin as he ends where he started was 0 m/s.

To solve more questions on kinematics, visit the link below-

brainly.com/question/27200847

#SPJ1

4 0
1 year ago
Microwave ovens use microwave radiation to heat food. the microwaves are absorbed by the water molecules in the food, which is t
PIT_PIT [208]
<span>step 1: energy required to heat coffee E = m Cp dT E = energy to heat coffee m = mass coffee = 225 mL x (0.997 g / mL) = 224g Cp = heat capacity of coffee = 4.184 J / gK dT = change in temp of coffee = 62.0 - 25.0 C = 37.0 C E = (224 g) x (4.184 J / gK) x (37.0 C) = 3.46x10^4 J step2: find energy of a single photon of the radiation E = hc / λ E = energy of the photon h = planck's constant = 6.626x10^-34 J s c = speed of light = 3.00x10^8 m/s λ = wavelength = 11.2 cm = 11.2 cm x (1m / 100 cm) = 0.112 m E = (6.626x10^-34 J s) x (3.00x10^8 m/s) / (0.112 m) = 1.77x10^-16 J step3: Number of photons 3.46x10^4 J x ( 1 photon / 1.77x10^-16 J) = 1.95x10^20 photons</span>
7 0
3 years ago
Read 2 more answers
A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje
kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0
3 years ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
Read 2 more answers
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