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strojnjashka [21]
3 years ago
11

1. A woman driving in a car at 21 m/s pushes down on the gas pedal. 6 seconds later, she is moving at 36 m/s. What was her avera

ge acceleration?
Physics
1 answer:
Furkat [3]3 years ago
5 0

a=Vf-Vi/t=36-21/6=15/6=2.5m/s^2

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F.r.e.e points :]]] ​
Colt1911 [192]
Ok? I don’t know what you want me to do though
4 0
2 years ago
Read 2 more answers
A 50-kg copper block initially at 140°c is dropped into an insulated tank that contains 90 l of water at 10°c. Determine the f
xxMikexx [17]

Answer:

T_f=24.71

Explanation:

From the question we are told that:

Mass of block m=50

Temperature of block T_b =140 \textdegree C

Volume of water V= 90L

Temperature of water T_w=10 \textdegree C

Density of water \rho=1000kg/m^3

Specific heat of water C_w=4.18KJ/kg-k

Specific heat of copper C_p=0.96KJ/kg-k

 

Generally the equation for equilibrium stage is mathematically given by

mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)

50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)

48(140-T_f)=376.2(T_f-10)

140-T_f=7.8375(T_f-10)

140-T_f=7.8375T_f-78.375

-8.8375T_f=-218.375

T_f=\frac{-218.375}{-8.8375}

T_f=\frac{-218.375}{-8.8375}

T_f=24.71

6 0
2 years ago
Consider the minute hand on a clock. (a) Compute the frequency of its motion in cycles per second. State your answer to three si
kicyunya [14]

Answer:

(a)0.0002778

(b)1.16\times10^{-5}

Explanation:

(a) The minute hand has a period of 60 minutes ( or 60 * 60 = 3600 seconds) for 1 circle. Its frequency per second would be

1 / 3600 = 0.0002778

(b) The hour hand has a period of 24 hours ( or 24*60 * 60 = 86400

seconds) for 1 circle. Its frequency per second would be

1 / 86400 = 1.16\times10^{-5}

5 0
2 years ago
Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the
mash [69]

Answer:

The value is }  N  =  66 \  blocks

Explanation:

From the question we are told that

The weight of the block is W_b  = 100 \  N

The dimension of the block is d =  0.400 m  \ X  \ 0.250 \  m  \  X  \ 0.130 \ m

Generally two atmosphere is equivalent to

P_{2atm} =  2 *  1.013 *10^{5} =  202600 \  N/m^2

Generally 1 atm = 1.013 *10^{5} N/m^2

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

A =  0.250 *  0.130

=> A =  0.0325 \  m^2

Generally the force due to this blocks is mathematically represented as

F =  N  *  W_b

Here N is the number of blocks

So

}  202600 =  \frac{N  *  100 }{ 0.0325}

=>   }  N  =  66 \  blocks

3 0
3 years ago
Is
ivann1987 [24]

Answer:

Required energy Q = 231 J

Explanation:

Given:

Specific heat of copper C = 0.385 J/g°C

Mass m = 20 g

ΔT = (50 - 20)°C = 30 °C

Find:

Required energy

Computation:

Q = mCΔT

Q = 20(0.385)(30)

Required energy Q = 231 J

4 0
2 years ago
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