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3 years ago

An object with more mass has more kinetic energy than an object with less mass, if both objects are moving ____.

2 answers:

6 0

An object with more mass has more kinetic energy than an object with less mass, if both objects are moving at the same speed. <em>(c)</em>

creativ13 [48]3 years ago

4 0

Answer:

c. at the same speed

Explanation:

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A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will

77julia77 [94]

Answer:

$V=200cm^3\\\\\rho =0.500g/cm^3$

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

$V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3$

Moreover, since the density is computed via the division of the mass by the volume:

$\rho =\frac{m}{V}$

We obtain:

$\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3$

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

3 0

3 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago

The area under the curve of the net external force vs time graph is equal to __________ or ________.

Sophie [7]

Impulse delivered

Change in momentum.

7 0

3 years ago

During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshi

Allushta [10]

Answer:

d = 0.38 m

Explanation:

As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:

vf = v₀ -a*t

If vf = 0, we can solve for v₀:

v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s

With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.

Just for simplicity, we can use the following equation:

$vf^{2} -vo^{2} = 2*a*d$