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cestrela7 [59]
3 years ago
5

A runner achieves a velocity of 11.1 m/s nine seconds after he begins. What is his acceleration

Physics
2 answers:
Delvig [45]3 years ago
6 0

Answer:

=1.23m/s^{2}  (2 decimal places

Explanation:

bogdanovich [222]3 years ago
6 0

Answer:

1.23

Explanation:

Because if we divide: 11.1 ÷ 9 = 1.23

to make sure: 1.23 x 9 = 11.07

and that's the closest to 11.1 so it count's as an answer.

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An airplane travels 4000m in 16 seconds on a heading of 35 whats is its velocity
Marizza181 [45]

In the question, you just gave a complete and detailed
description of the plane's velocity vector:

       4,000/16  meters/second , heading 35 degrees .

You might want to simplify the speed and make it a unit rate,
but otherwise, it's perfect.

         250 meters/second, heading 35 degrees .

6 0
3 years ago
The standard normal curve is not symmetrical. <br> a. True<br> b. False
FrozenT [24]

Answer;

The above statement is false

Explanation;

Symmetrical distribution, commonly known as symmetric distribution or normal distribution, is typically unimodal, meaning it shows only one peak in graph form.

It is a type of distribution where the left side of the distribution mirrors the right side. By definition, a symmetric distribution is never a skewed distribution.

All normal distributions are symmetric and have bell-shaped density curves with a single peak.

7 0
3 years ago
Read 2 more answers
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
Jill leaves home and rides a distance of 70 km. It took her 2.5 hours. What is her speed?
Paraphin [41]
The answer is ...
28 km per hour
5 0
3 years ago
An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl
mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0
3 years ago
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