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1 year ago

Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Physics

1 answer:

Andre45 [30]1 year ago

6 0

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

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Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ 3 2 4 ⎦ ⎥ ⎤ km away from the origin with respect to so

Anettt [7]

Answer:

[1, 6, -2]

Explanation:

Given the following :

Initial Position of spaceship : [3 2 4] km

Velocity of spaceship : [-1 2 - 3] km/hr

Location of ship after two hours have passed :

Distance moved by spaceship :

Velocity × time

[-1 2 -3] × 2 = [-2 4 -6]

Location of ship after two hours :

Initial position + distance moved

[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]

= [3-2, 2+4, 4-6] = [1, 6, -2]

4 0

2 years ago

Some car manufacturers claim that their vehicles could climb a slope of 42 ∘. For this to be possible, what must be the minimum

Paladinen [302]

Answer:

D. 0.9

Explanation:

Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.

Resolving into x component:

∑ $F_{x} = F_{s} - mgsin\alpha =0$

$F_{s} = mgsin\alpha$ ------(1)

Resolving into y component:

∑ $F_{y} = F_{n} - mgcos\alpha =0$

$F_{n} = mgcos\alpha$ ------(2)

Static frictional force, $F_{s} \leq$ μ $F_{n}$ ------(3)

substituting $F_{s}$ from equation (1) and $F_{n}$ from equation (2) into equation (3)

$mgsin\alpha \leq$ μ $mgcos\alpha$

$sin\alpha \leq$ μ $cos\alpha$

μ $\geq \frac {sin\alpha}{cos\alpha}$

μ $\geq tan\alpha$

The angle the vehicles make with the horizontal α = 42°

μ ≥ tan 42°

μ ≥ 0.9

6 0

1 year ago

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

olya-2409 [2.1K]

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

m=8/32=0.25

To find the spring constant

Kx=mg (given that c=6 in and mg=8 lb)

K(0.5)=8 (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0

2 years ago

A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

g100num [7]

Explanation:

The given data is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And, $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

Force = Pressure difference × A

= $(100000 - P) \times 2.82 \times 10^{-3}$

We are given that force is 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

PV = nRT

$38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

T = 462.66 K

Thus, we can conclude that temperature of the gas is 462.66 K.

6 0

2 years ago

A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme

GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

East component = 250 * cos 60°

East component = 125 km

8 0

2 years ago

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