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ki77a [65]
3 years ago
14

A place where things are baked ​

Physics
1 answer:
slava [35]3 years ago
4 0
“A place where things are baked”

- the bakery?
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Which of the following is a Nobel gas?<br> Ba2+<br> OA<br> OB. Ci<br> Kr<br> Ос.<br> Ca<br> D.
stepan [7]

Answer:

Kr

Explanation:

the element that is in group 18 is noble gasses. the elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

5 0
3 years ago
A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The s
guapka [62]

Answer:

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

Explanation:

Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.

In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.

Therefore, to test the validity of the claim, the student should use the following measuring tools;

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.

b. A meterstick to measure the distance of the track that the car travels on.

Hence, with these two devices the student can effectively measure or determine the validity of the claim.

5 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
PLEASE HELP QUICKLY 13 POINTS +7 FOR BEST ANSWER!!! calculate the density of an object that has a mass of 3.0 g and a volume of
raketka [301]
Density = 3/5 = 0.6g/cm^3. Since the density is less than the density of water, which is 1, the object will float.
8 0
3 years ago
Read 2 more answers
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

mg sin \theta - \mu N = ma (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, \theta=39^{\circ} being the angle of the incline

\mu N is the frictional force, with \mu = 0.6 being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

N-mg cos \theta =0

where mg cos \theta is the component of the weight in the direction perpendicular to the slope. Solving for N,

N=mg cos \theta

Substituting into (1), solving for a, we find the acceleration:

a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2

6 0
4 years ago
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