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Bas_tet [7]
2 years ago
5

A turtle starts at a velocity of 2 m/s and accelerates at a rate of 3 m/s2. How long will it take the turtle to reach a final ve

locity of 11 m/s?
Physics
1 answer:
luda_lava [24]2 years ago
6 0
U= 2m/s
a= 3m/s^2
v= 11m/s
t=?


We know that a= v-u/t
t = v-u/a
t = 11-2/3
= 9/3
=3 seconds


Hope it helps!
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The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the
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Compute first for the vertical motion, the formula is:

y = gt²/2 

0.810 m = (9.81 m/s²)(t)²/2 

t = 0.4064 s 


whereas the horizontal motion is computed by: 

x = (vx)t 

4.65 m = (vx)(0.4064 s) 

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed. 

(vy) = gt 

(vy) = (9.81 m/s²)(0.4064 s) 

(vy) = 3.99 m/s 


speed with which it hit the ground: 

v = sqrt[(vx)² + (vy)²] 

v = sqrt[(11.44 m/s)² + (3.99 m/s)²] 

v = 12.12 m / s
6 0
3 years ago
What is your speed (in m/s) if you walk 1000m in 20 minutes? (Hint: how many seconds are in a minute)
labwork [276]

Answer:

20min = 20 × 60 = 1200sec.

Speed in m per sec.

V = 1000/1200

V = 0.833m per sec.

Explanation:

7 0
3 years ago
A battery can provide a current of 1.80 A at 2.60 V for 6.00 hr. How much energy (in kJ) is produced?A battery can provide a cur
love history [14]

Answer:

The energy which is produced by a battery is 101.1 kJ.

Explanation:

The expression for the energy in terms of voltage, current and time is as follows;

E=VIt

Here, V is the voltage, I is the current and t is the time.

It is given in the problem that a battery can provide a current of 1.80 A at 2.60 V for 6.00 hr.

Calculate the energy of the battery.

E=VIt

Convert time from hour int seconds.

t=6 hr

t=(6)(60)(60)

t=21600 s

Put I= 1.80 A, V= 2.60 V and t= 21600 s in the expression of energy.

E=(2.60)(1.80)(21600)

E= 101.1 kJ

Therefore, the energy which is produced by a battery is 101.1 kJ.

4 0
3 years ago
An object needs a force of 152 Newtrons to move 8 meters. How much work is required?
GuDViN [60]

Answer:

960J

Explanation:

Given parameters:

Force  = 120N

Distance = 8m

Unknown:

Work required  = ?

Solution:

The work done by a body is the force applied to move a body in a specific direction.

 Work done  = Force x distance

Insert the parameters and solve;

  Work done  = 120  x  8  = 960J

6 0
3 years ago
The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh
stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, P_{headlight} = 38 W

The power at which the kick starter operates in parallel, P_{kick \ starter} = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/P_{headlight}  = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/P_{kick \ starter} = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

P_{headlight} = I² × R₁

∴ P_{headlight} = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, P_{headlight, S} ≈ 0.134 W

P_{kick \ starter} = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, P_{kick \ starter, S} ≈ 2.123 W

The total power they will consume, P_{Total} = P_{headlight, S} + P_{kick \ starter, S}

Therefore;

P_{Total} ≈ 0.134 W + 2.123 W = 2.257 W

4 0
3 years ago
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