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lbvjy [14]
3 years ago
8

A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22.0-min rest stop. If the person’s average spee

d is 77.8 km/h, (a) how much time is spent on the trip and (b) how far does the person travel?
Physics
1 answer:
DanielleElmas [232]3 years ago
7 0

Answer:

(a): It spends on the trip 2.75 hr (2 hours 45 minutes)

(b):  The person travels 213.9 km.

Explanation:

Vp= 77.8 km/h

V1= 89.5 km/h

V2= 0 km/h

t2= 0.36hr

t1= ?

Vp= (V1*t1 + V2*t2) / (t1 + t2)

clearing t1:

t1= 2.39 hr

Trip time= t1+t2

Trip time= 2.75 hr (a)

Distance traveled= V1*t1

Distace traveled= 213.9 km (b)

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Artemon [7]

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

I = mK^{2}

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

6 0
3 years ago
I need help with the table, which is speed and velocity and acceleration
likoan [24]
The first box is Acceleration
the second box is Speed
the third box is Speed
the fourth box is velocity
the fifth box is Speed.

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The reason the second, third,and fifth box is speed is because it is telling up that an object is going _mph (the blank is where you put your estimate of speed of an object). 

The reason the fourth box is velocity is because velocity is both magnitude and direction so in the fourth box it says that the race car made a slight left turn which is a direction that the car is going in. 

Hope this helped :)
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5 0
3 years ago
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Which force is a field force?
jok3333 [9.3K]
it has to me c because it’s a part of the force of things that do
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A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave
Evgesh-ka [11]

Answer:

34.3 m/s

Explanation:

We can consider the car as it is in a vertical circular motion. There are two forces acting on the car:

- The weight of the car: mg, where m is the mass of the car and g is the gravitational acceleration, acting downward

- The normal force exerted by the road on the car: N, acting upwards

Using Newton's second law, we can write that the resultant of these two forces must be equal to the product between the mass of the car, m, and the centripetal acceleration, a_c:

mg-N=ma_c = m\frac{v^2}{r}

where v is the speed of the car and r = 0.120 km = 120 m is the radius of the circular trajectory.

We are asked to find the speed of the car when its tires just barely lose contact with the road: this means when the normal force exerted by the road on the car becomes zero, N=0. Substituting this information into the previous equation and solving for v, we find the speed of the car:

mg=m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s

3 0
3 years ago
Charge of uniform density (40 pC/m^2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical
Nataly [62]

Answer:

The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C

Explanation:

Step 1: Data given

Charge of uniform density of first sphere= 40 pC/m² = 40.0 * 10^-12 C/m²

Radius of first sphere is 1.0 cm

Radius of second sphere = 3.0 cm

Charge of uniform density of second sphere= 60 pC/m² = 60.0 * 10^-12 C/m²

Step 2: Calculate the magnitude

Sphere surface area = 4πr²

Charge on inner sphere Qi = 40.0*10^-12C/m² * 4π(0.01m)² = 5.027*10^-14 C

NET charge on outer sphere Qo = 60.0*10^-12 * 4π(0.03m)² = 6.786*10^-13 C

Inner sphere induces a - 5.027*10^-14 C  charge (-Qi) on inside of the outer shell

This means there is a net zero charge within the outer shell.

For the outer shell to show a NET charge +6.786*10^-13C, it's must have a positivie charge {+6.786*10^-13C + (+5.027*10^-14C)} =  +7.2887*10^-13 C

Regarding the outer shell as a point charge (field at 0.04m is)

E = kQ /r²

E = (8.99^9)*(7.2887*10^-13 C) / (0.04)² .. .. ►E = 4.10 N/C

========

Using Gauss's law .. with a spherical Gaussian surface  at 0.04m enclosing a net charge +7.2887*10^-13 C

Flux EA = Q / εₒ

  ⇒ with εₒ  = 8.85 *10^-12

⇒ with Q = 7.2887*10^-13 C

E * 4π(0.04)² = 7.2887*10^-13 C / 8.85*10^-12  

E =  7.2887*10^-13 C / {8.85*10^-12 * 4π(0.04)²}

E = 4.10 N/C

The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C

3 0
3 years ago
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