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jekas [21]
4 years ago
12

As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the

bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.
Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?
Physics
1 answer:
Lesechka [4]4 years ago
8 0

Answer:

a.5.66ms^{-2}

b.55 m

Explanation:

We are given that

Mass ,m=0.9 kg

Length of string,l=50 cm=\frac{50}{100}=0.50 m

1 m=100 cm

\theta=30^{\circ}

R=55 m

a.Centripetal acceleration

a_c=gtan\theta

a_c=9.8tan30^{\circ}

a_c=5.66 m/s^2

Hence, the radial acceleration of the bus=5.66m/s^2

b. Radius of curve,R=55 m

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expeople1 [14]

Answer: X = 52,314.12 N

Explanation: Let X be the force the feet of the athlete exerts on the floor.

According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)

Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.

The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s

The acceleration of the barbell is gotten by using the equation of constant acceleration motion

S= ut + 1/2at²

But u = 0

S = 1/2at²

0.65 = 1/2 *a (1.3)²

0.65 = 1.69 * a/2

0.65 * 2 = 1.69 * a

a = 0.65 * 2/ 1.69

a = 0.77m/s²

According to newton's second law of motion

Resultant force = mass * acceleration

And resultant force in this case is

X - 52,200 = (87 + 61.22) * 0.77

X - 52,200 = 148.22 * 0.77

X - 52, 200 = 114.132

X = 114.132 + 52,200

X = 52,314.12 N

6 0
3 years ago
Which characteristic does an object with a constant acceleration always have?
Alexus [3.1K]

By definition, speed is the integral of acceleration with respect to time.

We have then:

v = \int\limits^t_0 {a} \, dt

As the acceleration is constant, then integrating we have:

v = a*t + vo

Where,

vo: constant of integration that corresponds to the initial velocity

We observe then that the speed varies linearly when the acceleration is constant .

Therefore, for constant acceleration, the velocity is changing.

Answer:

an object with a constant acceleration always have:

A. changing velocity

6 0
3 years ago
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3 years ago
Should your parent Ignore you? Should your parent call you names?<br> Why/ Why not? (5pts)
AysviL [449]

no cause kids will think its Neglect and names will hert the kids and think they are not wanted

4 0
4 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
4 years ago
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