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Damm [24]
3 years ago
12

a. How high did the car climb up the hill after bouncing? Did it go higher lower or the same height as the drop position?

Physics
1 answer:
weeeeeb [17]3 years ago
8 0
THe answer is the car went higher
You might be interested in
If you heat a fixed quantity of gas, which of the following statements are true?Check all that apply.a.The volume will always in
kherson [118]

Answer:

Explanation:

Given

there is fix Quantity of gas i.e. mass of gas is constant

From ideal gas Equation

PV=nRT

(a) volume always increases is not true as Pressure can also be increased .

(b)If Pressure is constant along with mass then as Temperature increases, Volume also increases.

(c)true , Product of Pressure and volume depends upon temperature thus it also increases with temperature.

(d)Density of gas may or may not increases

As density is \frac{mass}{volume}

volume may increase or decrease as temperature increase .

(e)false

as it clearly stated that quantity of fixed therefore there is no change in gas

8 0
3 years ago
HELPPPPP !!!!
Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0
2 years ago
A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu
xeze [42]

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

8 0
2 years ago
A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0
3 years ago
SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
2 years ago
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