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2 years ago

The diagram shows a position-time graph What is the displacement of the object

Physics

1 answer:

algol132 years ago

5 0

The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.

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Eduardwww [97]

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5 0

3 years ago

What force is needed to give a 0.25-kg arrow an acceleration of 196 m/s/s

kakasveta [241]

Force (f) = ?

Acceleration (a) = 196 m/s^2

Mass (m) = 0.25 kg

F = (m) • (a)

F = (0.25) • (196)

F = 49 N

Answer : 49 N

I hope that helps you!! Any more questions??

8 0

3 years ago

Read 2 more answers

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

2 years ago

A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115= 3.175

now,

$a= \sqrt{a_t^2+a_r^2}$

$a= \sqrt{1.47^2+3.175^2}$

a= 3.49 m/s^2

3 0

3 years ago

A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide

11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed u = 2.00 m/s

Solution

Work = Change in kinetic energy

$F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s$

7 0

2 years ago