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3 years ago

Helpppp mee

2 answers:

7 0

Distance is the total length of the follwed by a movig partical.
Velocity is the rate of change of distance with respect to time. acceleration is the rate of change of velocity with respect to time.

RideAnS [48]3 years ago

6 0

Distance is measurable velocity is the speed of something and acceleration is the capacity to gain speed within a short amout of time
EXAMPLES
distance how far utah is from california
velocity the time it take you to run home
acceleration is like how soon a car reaches 25 mph

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Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

$V2=\frac {M1v1}{m2}$

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

$V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s$

4 0

3 years ago

suppose that a large cargo truck needs to cross a bridge. the truck is 30 m long and 3.2 m wide. the cargo exerts a force nof 54

Tamiku [17]

<span>No, because the truck applies more pressure than the bridge can support.</span>

7 0

3 years ago

The variation in the pressure of helium gas, measured from its equilibrium value, is given by ΔP = 2.9 × 10−5 cos (6.20x − 3 000

nadya68 [22]

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

$\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)$ ..............(1)

The general equation is given by :

$\Delat P=P_o\ cos(kx-\omega t)$ ...........(2)

On comparing equation (1) and (2) :

$k=6.2$

Since, $k=\dfrac{2\pi}{\lambda}$

$\dfrac{2\pi}{\lambda}=6.2$

$\lambda=1.01\ m$

So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.

6 0

3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago

VELOCITY, SPEED, DISTANCE ETC..

vaieri [72.5K]

Answer:

Explanation:

Speed = distance / time

Velocity = displacement / time

So ,

Speed = 50 km / 0.5 hr = 100 km/h

Velocity = 40 km / 0.5hr = 80 km/h

4 0

3 years ago