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ZanzabumX [31]
3 years ago
12

A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy

is closer to the center. Who has greater tangential acceleration?Both the girl and boy have the same nonzero tangential accelerationThe boy has greater tangential accelerationBoth the girl and the boy have zero tangential accelerationThe girl has greater tangential acceleration
Physics
1 answer:
galina1969 [7]3 years ago
5 0

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration (\alpha) of the merry go round is equal to the rate of the change of the angular velocity, \omega:

\alpha = \frac{d\omega}{dt}

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

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2. How much do you think the force of friction must be? Why?
Tamiku [17]

Answer:

It must be high do to the gravity

Explanation:

4 0
3 years ago
A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo
DIA [1.3K]

Answer:

74^{\circ} C

Explanation:

We are given that

Mass of glass,m=300 g

T_1=23^{\circ}

Volume,V=236cm^3

Mass of water=density\times volume=1\times 236=236 g

Density of water=1g/cm^3

Temperature of hot water,T=87^{\circ}

Specific heat of glass,C_g=0.2cal/g^{\circ}C

Specific heat of water,C_w=1 cal/g^{\circ}C

Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)

Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)

Q_{glass}+Q_{water}=0

300\times 0.2(T_f-23)+236\times 1(T_f-87)

60T_f-1380+236T_f-20532=0

296T_f=20532+1380=21912

T_f=\frac{21912}{296}=74^{\circ} C

5 0
3 years ago
A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f
Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

Acceleration due to gravity is, g=9.8\ m/s^2

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N

Therefore, the normal force acting on the dog is 210 N.

6 0
4 years ago
Please hurry
viktelen [127]

Answer:

The velocity will be "76.8 m/s".

Explanation:

The given values are:

Acceleration,

a = 2.4 m/s²

Time,

t = 32 seconds

By equation of motion,

⇒  v=u+at

On substituting the values, we get

⇒     =0+2.4\times 32

⇒     =0+76.8

⇒     =76.8 \ m/s

6 0
3 years ago
Help ASAP plsssss:///
kenny6666 [7]

Answer:

c

Explanation:

8 0
3 years ago
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