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Mandarinka [93]

1 year ago

6

Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s.

What is the acceleration?

1 answer:

Nataliya [291]1 year ago

7 0

Answer:

bonita sisisisiisisisisiisissiissiisiiss

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Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

Temka [501]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

6 0

1 year ago

Read 2 more answers

Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerate

____ [38]

Answer:

a) Velocity = 4.2m/s

b) Acceleration = 2.94m/s^2

c) Force exerted on the floor= 1401.4×10^3N

Explanation:

a) Velocity,V=sqrt(2×9.8×0.900)

V= 4.2m/s

b) Vf2= V^2+2ay2

a= 4.2^2 - 0/2×3

a= 17.64/6= 2.94m/s^2

c) Newton's 2nd law indicates:

Fnet= F - mg=ma

F= m(g+a)

F=110(9.8+2.94)

F=110×12.94

F= 1401.4N

4 0

1 year ago

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

= $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

= $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

ma = qE

a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

= $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0

1 year ago

At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m

Margaret [11]

Answer:

Part a)

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = \sqrt{\frac{2h}{g}}$

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

$d = 2.86 m$

so the distance from the edge of the pool is given as

$L = 2.86 + 2$

$L = 4.86 m$

4 0

1 year ago

How many turns should a 10-cm long ideal solenoid have if it is to generate a 1.5-mT magnetic field when 1.0 A of current runs t

Ira Lisetskai [31]

Answer:

N=119.34 turns

Explanation:

The magnetic field of a solenoid is calculated using the formula:

B= µo* $\frac{I*N}{L}$ Equation 1

Where:

B: magnetic field in Teslas (T)

µo: free space permeability in T*m/A

I= Intensity of the current flowing through the conductor in ampere (A)

N= number of turns

L= solenoid length in meters (m)

Data of the problem:

L=10cm= $10*10^{-2}$ , B= 1.5mT= $1.5*10^{-3} T$ ,I=1A

µo= $4\pi *10^{-7} \frac{T*m}{A}$

We cleared N of the equation (1):

N=B*L/ µo*I

N= $(1.5*10^{-3} *10*10^{-2} )/(4\pi *10^{-7} *1$

$N=0.1193*10^{3}$

Answer

N=119.34 turns

3 0

2 years ago