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oksian1 [2.3K]
2 years ago
10

Brad is working on a speed problem in physics class. The problem tells him that a girl runs from her house to the park 0.05 km a

way in 10 s. Brad calculates that her speed is 0.005 m/s. Is he correct? If not, explain the flaw or flaws in his problem solving process.
Physics
2 answers:
Dmitriy789 [7]2 years ago
7 0
It is corecct becasue it 10 s more added from 0.05
Olegator [25]2 years ago
3 0

Answer:

He is incorrect!  Her speed was 5m/s.

Explanation:

For calculating the speed, first we shall remember that:

v=\dfrac{d}{t}

Where v is the speed, d is the distance travelled and, t is the time it takes to travel distance d.

So one migth think that velocity can be easely compute:

v=\dfrac{0.05}{10}

v=0.005\dfrac{m}{s}

Be carefull, he does not make a proper dimensional analisis!  

Before computing the speed we must know in what dimensions our values are.

d=0.05km, distances is measure in Kilometers.

t=10s, time is measure in seconds.

If we want our speed to be in m/s, first we need to be sure that our values are expressed  in meters and seconds.

Time is already  expressed in seconds, distance is not in Kilometers.

So

0.05Km=50m,

now we can compute the speed:

v=\dfrac{d}{t}

v=\dfrac{50m}{10s}

v=\5dfrac{m}{s}

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Imagine you want to get 1 kcal of energy from a cow. How much energy would the cow need to get from plants? Why?
ZanzabumX [31]
1000 kcal because you only get 10% of the energy of the thing you eat
7 0
2 years ago
A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a
Korvikt [17]

Answer:

v^2=v_o^2-2\times a\times (v_o.t)

Explanation:

Given:

Initial velocity of the vehicle, v_o

distance between the car and the tree, x_f

time taken to respond to the situation, t

acceleration of the car after braking, a

Using equation of motion:

v^2=u^2+2a.s ..............(1)

where:

v= final velocity of the car when it hits the tree

u= initial velocity of the  car when the tree falls

a= acceleration after the brakes are applied

s= distance between the tree and the car after the brakes are applied.

s=v_o\times t

Now for this situation the eq. (1) becomes:

v^2=v_o^2-2\times a\times (v_o.t) (negative sign is for the deceleration after the brake is applied to the car.)

5 0
2 years ago
URGENT!!!!
Nastasia [14]

Answer:

8 cm

Ladder is safer than to walk on it

Explanation:

Streched = Final length - initial length

= 66 - 58

= 8 cm

When your are walking along frozen ice pond your weight is not distributed .It's acting on directly on the ice pond because of contact Area is too small between pond and you. There for According to pressure equation ,

P = F/A

When area is less pressure is less.

If use ladder instead of going by foot. Contact area will be high between pond and ladder. There for pressure is reduced .

3 0
2 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
2 years ago
Based on the emf value measured at frame 700, what is the average magnitude of the magnetic field inside the magnet assembly? No
Nadusha1986 [10]

Answer:

The average magnitude of magnetic field B= 0.0433/ d Tesla

(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)

Explanation:

Induced emf

where B= magnetic field  

d= breadth of rectangular piece

V= velocity with which the rectangular piece = o.o6m/s

n= no of turns  = 10

EMF = 26mV

since d (breadth of the frame) is not given, I will use it as a variable

EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)

From eq 1, we get

B= (EMF)/(n d V)

B= (26 X 0.001) / (10 d 0.06)

B= 0.0433/ d Tesla

4 0
2 years ago
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