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Trava [24]
2 years ago
8

To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a

ssuming that the tea was initially at 72f, what is the final temperature of the tea in f?
a. 33
b. 91
c. 110
d. 140
e. 200
Physics
1 answer:
IRISSAK [1]2 years ago
4 0
Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91. 
The answer in this question is B. 91
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A normal polarity magnet moves toward a stationary coil at 20 cm/s, and induces a maximum current of –8 mA. Which scenarios woul
maksim [4K]

<em>If the distance between the two objects is the same, then;</em>

Both the magnet and the coil moving toward each other at 10 cm/s each

A reversed polarity magnet moving away from the coil at 20 cm/s

<u>Calculate current that produces a magnetic field, and use the right hand rule 2, to determine the direction of current or the direction of magnetic field loops. </u>

6 0
2 years ago
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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration
wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

V=X/T

First stage:

T1=5s

v_{f}  =v_{o} - at

But, v_{f}  =0   (decelerates to rest)

then: a =v_{o} /t=0.3/5=0.06m/s^{2}

on the other hand:

x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m

X1=75cm

Second stage:

T2=5s

x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0
2 years ago
Karen is running forward at a speed of 9 m/s. She tosses her sweaty headband backward at a speed of 20 m/s. The speed of the hea
Komok [63]
Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s

The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s

The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s

Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.

8 0
2 years ago
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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

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-i</span>
3 0
2 years ago
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You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for
mart [117]

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

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= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

8 0
2 years ago
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