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3 years ago

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A ferry boat is 4.0 m wide and 6.0 m long. when a truck pulls onto it, the boat sinks 4.00 cm in the water. what is the weight o

f the truck?

1 answer:

just olya [345]3 years ago

7 0

Answer : Weight of the truck = 9408 N

Explanation :

Given that,

Length = 6.0 m

Width = 4.0 m

Depth = 4.00 cm

we know that,

weight of the truck = weight of the displaced water

weight of the truck = $m_{water}\times g$

we know that,

density $\rho = \dfrac{mass}{volume}$

mass = density X volume

weight of the truck = $\rho_{w}\times V_{w}\times g$

Now, the volume = length X width X depth

$V = 6.0\ m\times 4.0\ m \times 0.04\ m$

$V = 0.96\ m^{3}$

Weight of the truck $= 1000\ kg/m^{3}\times 0.96\ m^{3}\times 9.8\ m/s^{2}$

Weight of the truck = $= 9408\ kg- m/s^{2}$

Weight of the truck = 9408 N

Hence, this is the required solution.

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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

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Your backpack has a mass of 8 kg. You drop it from a height of 1.3m. How much work is done by gravity as the backpack falls?

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Answer:

The answer is C.

Explanation:

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2 years ago

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A 0.80 kg object tied to the end of a 2.0 m string swings as a pendulum. At the lowest point of its swing, the object has a kine

devlian [24]

Answer:

3.32 m/s

Explanation:

From the law of conservation of energy, the sum of mechanical and kinetic energy should be equal to the 10 J given. Potential energy is given by mgh where m is mass, g is acceleration due to gravity and h is the height. For this case, $h= l(1-cos\theta)$ and l is string length, given as 2 m, \theta is given as 50 degrees. Kinetic energy is given by $0.5mv^{2}$ and it is this velocity that is unknown.

$10J=0.5\times 0.8kg\times v^{2}+ 0.8kg\times 9.81\times 2m(1-cos 50^{\circ})\\v\approx 3.32 m/s$

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3 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

So

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

Since the value is positive it implies that the cart moved eastward

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3 years ago