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4 years ago

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A grocery shopper tosses a(n) 9.0 kg bag of rice into a stationary 17.4 kg grocery cart. The bag hits the cart with a horizontal

speed of 5.2 m/s toward the front of the cart. What is the final speed of the cart and bag

1 answer:

noname [10]4 years ago

7 0

Answer:

$V=1.77m/s$

Explanation:

#Using the conservation of momentum , momentum before equals momentum after( $p=mv)$ .

-Initial speed is 5.2m/s while the cart is at rest. After, the velocity will be of a combined(bag+cart) mass.

Hence:

$9.0\times 5.2+17.4\times 0=(9+17.4)V\\\\V=\frac{9\times5.2}{9+17.4}\\\\=1.77m/s\\$

Hence, the final velocity of the cart and bag is 1.77m/s

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A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i

konstantin123 [22]

Answer:

$T_f=5854.76$ °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, $E= 1.10\times 10^6 J$

heat capacity of the hiker, $c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}$

initial body temperature of hiker, $T_i=36.6 \degree C$

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

$W=63\times 9.8\times 828$

W= 511207.2 J

∵Hike eats 2 energy bars= $2\times 1.1\times 10^{6} J$

Energy produced $= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

$25\% of E= 511207.2$

$\Rightarrow E= 511207.2\times \frac{100}{25}$

$E=2044828.8 J$

&

Amount of total energy (E) converted into heat(Q) is:

$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$ .................(1)

where:

$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

$T_f\approx 5854.76$ °C

4 0

3 years ago

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$

The maximum force is when the car is accelerating so

$Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N$

so the maximum force is the maximum force by the maximum speed

$Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW$

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

4 0

4 years ago

A series circuit has a 100-Ω resistor, 4.00-mH inductor and a 0.100-μF capacitor connected across a 120-V rms ac source at the r

scZoUnD [109]

Answer:

144 watt

Explanation:

resistance, R = 100 ohm

L = 4 mH

C = 100 micro farad

At resonance, the impedance is equal to R

Z = R

Vrms = 120 V

Irms = Vrms / R = 120 / 100 = 1.2 A

Power is given by

P = Vrms x Irms x CosФ

Where, CosФ is called power factor

At resonance, CosФ = 1

Power, P = Irms x Vrms

P = 1.2 x 120 x 1

P = 144 Watt.

Thus, the power is 144 watt.

6 0

4 years ago

SI units are used for the scientific works,why?

a_sh-v [17]

Answer:

SI is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion

8 0

3 years ago

A wire in a circuit carries a current of 0.9 A. Calculate the quantity of

raketka [301]

Answer:

We conclude that the quantity of the charge that flows through the wire in 50 s will be 45 C.

Explanation:

Given

Current I = 0.9 A

Time t = 50 s

To determine

We need to find the quantity of the charge that flows through the wire in 50 s.

Important Tip:

A current of 1 Ampere = 1 Coulomb of charge flowing in 1 second

Using the formula involving charge and current

$I=\frac{Q}{t}$

where:

$I$ represents the current in amperes (A)

$Q$ represents the charge in coulomb (C)

$t$ represents the time in seconds (s)

now substituting I = 0.9 and t = 50 in the formula

$I=\frac{Q}{t}$

$0.9\:=\:\frac{Q}{50}$

switch sides

$\frac{Q}{50}=0.9$

Multiply both sides by 50

$\frac{50Q}{50}=0.9\cdot \:50$

Simplify

$Q=45$ C

Therefore, we conclude that the quantity of the charge that flows through the wire in 50 s will be 45 C.

6 0

3 years ago