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KatRina [158]
3 years ago
9

the scores of players on a golf team are shown in the table. the teams combined score was 0 what was travis's score?

Physics
2 answers:
barxatty [35]3 years ago
7 0

Answer:

-5

Explanation:

Alona [7]3 years ago
3 0

Answer:

what table?

Explanation:

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PLEASE HURRY I HAVE A TIME LIMIT
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Properties of electromagnetic radiation and photons. ... we find the types of energy that are lower in frequency ( and thus longer in wavelength) than visible light. Seeee
4 0
3 years ago
. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?
Hoochie [10]

The acceleration of the object is 6.7 m/s^2

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
2 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
Read 2 more answers
What relationship exists between the amplitude of a wave and amount of disturbance in the water?
guapka [62]
A wave is a result of the disturbance in the equilibrium state.  There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.
6 0
3 years ago
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m

8 0
2 years ago
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