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34kurt
3 years ago
13

A 3kg object moving at 4 m/s encounters a 20 N resistive force over a duration of 0.20 seconds. The impulse experienced by this

object is _____
Physics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

the impulse experienced by this object is 4 Ns

Explanation:

Given;

mass of the object, m = 3 kg

velocity of the object, v = 4 m/s

resistive force, F = 20 N

duration of impact, t = 0.2 s

The impulse experienced by this object is calculated as follows;

J = F x t

J = 20 x 0.2

J = 4 Ns

Therefore, the impulse experienced by this object is 4 Ns

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A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t
Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, \epsilon=1.5\ V

Magnitude of Earth's field, B=10^{-4}\ T

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}

\omega=159.15\ rad/s

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0
3 years ago
An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =
Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V_m = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V_m =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

I(t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

I(t) = V₀√(C/L)

we substitute

I(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

I(t) = 25 × 0.00002

I(t) = 0.0005 A

I(t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0
3 years ago
A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
3 years ago
A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

3 0
3 years ago
What is a primary source?
Fed [463]
//////Correct answer is C.///////
6 0
3 years ago
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