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Vitek1552 [10]
3 years ago
12

The free body diagram represents Silly Sally hanging from a trapeze bar. Sally weighs 660 Newtons. What is the force in each of

the chains holding the trapeze bar?
A. 1320 N

B. 185 N

C. 660 N

D. 330 N

Physics
1 answer:
MrRissso [65]3 years ago
5 0

the force in each of the chains holding the trapeze bar is D. 330 N

Her weight has to be divided evenly so, divide 660/2= 330 N

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When steam condenses 1. All of these occur. 2. None of these occur. 3. molecules move closer together. 4. it changes from the ga
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What is the momentum of a 200 kg truck travelling at 20 m/s?
geniusboy [140]

Answer:

p = 4000 kg-m/s

Explanation:

Given that,

The mass of a truck, m = 200 kg

Speed of the truck, v = 20 m/s

We need to find the momentum of the truck. The formula for momentum is given by :

p = mv

so,

p=200\times 20\\\\p=4000\ kg-m/s

So, the momentum of the truck is equal to 4000 kg-m/s.

8 0
3 years ago
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

5 0
3 years ago
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