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AleksAgata [21]
3 years ago
9

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2

.0 m/s2 at the instant when David passes. How far does Tina drive before passing David?

Physics
1 answer:
dangina [55]3 years ago
5 0
A. 441 m B: 46.0 m/s
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1. A baseball is thrown vertically at 16.7 m/s. What is the maimum height of the baseball?
Anvisha [2.4K]

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

7 0
4 years ago
The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in
tangare [24]
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
5 0
3 years ago
Necesito ayudaaaaaa por favor
natima [27]

MAnswer:

Explanation:

4 0
3 years ago
Read 2 more answers
Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20
timama [110]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0
3 years ago
Read 2 more answers
1.A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam c
SIZIF [17.4K]

Answer:

the power that can be generated by the river is 117.6 MW

Explanation:

Given that;

Volume flow rate of river v = 240 m³/s

Height above the lake surface a h = 50 m

Amount of power can be generated from this river water after the dam is filled = ?

Now the collected water in the dam contains potential energy which is used for the power generation,

hence, total mechanical energy is due to potential energy alone.

E_{mech} = m(gh)

first we determine the mass flow rate of the fluid m

m = p×v

where p is density ( 1000 kg/m³

so we substitute

m = 1000kg/m³ × 240 m³/s

m = 240000 kg/s

so we plug in our values into ( E_{mech} = m(gh) kJ/kg )

E_{mech} = 240000 × 9.8 × 50

E_{mech} = 117600000 W

E_{mech} = 117.6 MW

Therefore, the power that can be generated by the river is 117.6 MW

4 0
3 years ago
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