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2 years ago

In Fig. on the right, what is the acceleration at 1.0 s?

Physics

2 answers:

AleksAgata [21]2 years ago

7 0

Answer:

10m/s2

Explanation:

Acceleration is defined as change in velocity over the time period

Now at 1second the velocity is 10m/s.

Hence the acceleration is V-U/t;

Where V is velocity at 10s and U is velocity when the object is at rest.

So we have;

Acceleration = 10-0/ 1 = 10m/s2

maks197457 [2]2 years ago

3 0

Answer:

10 m/s²

Explanation:

Acceleration is the slope of the velocity vs time graph.

At t = 1.0 s, the slope of the line is:

a = Δv/Δt

a = (20 m/s) / (2.0 s)

a = 10 m/s²

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How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?

Naddika [18.5K]

The amount of heat needed to raise the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
Cs is its specific heat capacity
$\Delta T$ is the increase in temperature

For oxygen, the specific heat capacity is approximately
$C_s = 0.92 J/(g K)$
The variation of temperature for the sample in our problem is
$\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K$
while the mass is m=150 g, so the amount of heat needed is
$Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J$

4 0

3 years ago

Read 2 more answers

Hello, how to do qn 5?:)l

Hitman42 [59]

Answer:

Explanation:

Please see the attached picture for the full solution.

Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.

7 0

2 years ago

PLZ HELP WILL GIVE BRAINLIEST

xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height = 8.18m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we use one of the motion equations.

v² = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

v² = 0² + (2 x 9.8 x 8.18)

v² = 160.3

v = 12.7m/s

7 0

3 years ago

Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

6 0

3 years ago

Plz answer brainliest if right!

AleksandrR [38]

Answer: it is D. it is the only possible answer. use the process of elimination. which answers make sense?

4 0

2 years ago