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IRISSAK [1]
2 years ago
3

A child on a spinning ride at a playground has a centripetal acceleration of 0.80 m/s2. The child completes a full circle every

4.2 s.
How far from the center of the ride is the child?

0.068 m
0.085 m
0.27 m
0.36 m
Physics
2 answers:
Novosadov [1.4K]2 years ago
5 0
The formula for centripetal acceleration is:
a = (4π²R)/T²
R = (aT²)/4π²
R = (0.8 x 4.2²) / 4π²
R = 0.36 m

The child is 0.36 meters away from the center.
Alenkasestr [34]2 years ago
5 0

Answer:

The answer is D

Explanation:

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What is the momentum of a 533 kg blimp moving east at +75 m/s
mylen [45]

Answer:

39975kgm/s due east

Explanation:

Given parameters:

Mass of the blimp  = 533kg

Velocity  = +75m/s due east

Unknown:

Momentum of the body  = ?

Solution:

The momentum of a body is the amount of motion it posses.

 Momentum is the product of mass and velocity;

 

  Momentum = mass x velocity

  Insert the parameters and solve;

    Momentum  = 533 x 75  = 39975kgm/s

The momentum of the body is 39975kgm/s due east

7 0
1 year ago
When compared to others, how is a greater velocity represented on a motion map?
kakasveta [241]
Motion map has the points spaced farther apart (because the car would go a further distance in each second), and the velocity vectors (arrows) are longer, because the car is moving faster. So 'with longer vectors' is the correct answer
8 0
2 years ago
Read 2 more answers
A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.
mafiozo [28]

Answer:

a. f=1.22*10^{-15} N

b. f=53.6*10^{-17} N

Explanation:

The force existing between two charges is given as

f=\frac{kq_{1}q_{2}}{r^{2}}

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

-1.602*10^{-19}

A. we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\  f=1.22*10^{-15} N

b. at  distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\  f=53.6*10^{-17} N

8 0
2 years ago
A piece of copper of mass 100 g is being drilled through with a 1/2" electric drill. The drill operates at 40.0 W and takes 30.0
lesantik [10]

Answer:

correct option is c. 31.0°C  

Explanation:

given data

copper of mass = 100 g

electric drill = 1/2"

power = 40.0 W

time = 30 s

C copper = 387 J/kgC

to find out

copper's increase in temperature

solution

we get here energy that is express a s

energy = 40 W × 30 s

energy = 1200 Watt seconds

and heat acquired by drill is here as

heat acquired = 100 × T × 387  

here temperature rise in copper mass as

temperature rise in copper mass = \frac{100}{1000} × T × 387

temperature rise in copper mass = 38.7 ×  T Watt second

we know that all the energy from the drill heats the copper

so we can say

38.7 ×  T = 1200

T = 31°C  

so correct option is c. 31.0°C  

4 0
2 years ago
A field researcher uses the slow-motion feature on her phones camera to shoot a video of an eel spinning at its maximum rate. Th
posledela

Answer:

\alpha=(1/120)*14*360=42degrees

Explanation:

Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way

So the angle the eel rotate is going to be

Camera record 120 frames per second

time taken for 1 frame record

t_t= 1/120 seconds

eel rotates

\alpha_1=14rev *360degrees in 1 second

so:

eel rotetas \alpha=(1/120)*14*360=42degrees

5 0
2 years ago
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