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3 years ago

A skier accelerates down the hill at 3m/s2 how fast is he going in 4 seconds

Physics

1 answer:

almond37 [142]3 years ago

7 0

Answer:

Explanation: simple kinematics

we suppose that initially vo= 0 so if the skier moves 4s :

vf = vo +at = 0 + 3*4 = 12 m/s

best wishes from colombia

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A spherical drop of water carrying a charge of 38 pC has a potential of 710 V at its surface (with V = 0 at infinity). (a) What

Hitman42 [59]

Answer: a) r= 4.82 * 10^-4 m ; b) 1420 V

Explanation: In order to solve this problem we have to take into account that potential for a sphere respec to V=0 at the infinity, which is given by:

V=k*Q/r where r is the radius of the drop

then we have

r=k*Q/V=9*10^9*38pC/710V= 4.82 * 10^-4 m

Finally if we join two drop to form one with the same radius but with twice charge the resultant potential is:

V= k*2*Q/(r)= 710*2= 1420 V

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4 years ago

Say you have a differential drive robot that has an axle length of 30cm and wheel diameter of 10cm. Find the angular velocity fo

Klio2033 [76]

Answer:

a) ω1 = 18rpm ω2 = -18rpm

b) ω1 = 102rpm ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm This is the angular velocity of the robot

$\omega = \frac{\omega r * D/2}{r_{wheel}}$ where D is 30cm and rwheel is 5cm

$\omega = \frac{6 * 30/2}{5}=18rpm$ One velocity will be positive and the other will be negative:

ω1 = 18rpm ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

$\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}$

$\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}$

Replacing values, we get:

$\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm$

$\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm$

For part c, both wheels must have the same velocity:

$\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min$

$\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm$

8 0

4 years ago

The circuit below represents four resistors connected to a 12-volt source. What is the total current in the circuit? 4.0Ω 6.0Ω 1

Tpy6a [65]

Answer:

For series connected resistors I = 0.5A

For parallel connected resistors I = 8.5A

Explanation:

Since the diagram is not available, our solution will be divided into two;

According to ohm's law which states that "the current passing through a,metallic conductor at constant temperature is directly proportional to the potential difference across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the total current in the circuit

Rt is total equivalent resistance.

Where E = 12volts

Rt can be gotten depending on the arrangements of the resistors which can either be in series or parallel.

If the resistors are in series, their equivalent resistance gives;

Rt = 4.0Ω+6Ω+8Ω+6Ω

Rt = 24Ω

The current I will be;

I = E/Rt = 12/24

I = 0.5A

If the connection is in series, the total current in the circuit will be 0.5A.

For resistance in parallel;

1/Rt = 1/4Ω+1/6Ω+1/8Ω+1/6Ω

1/Rt = 6+4+3+4/24

1/Rt = 17/24

Rt = 24/17Ω

I = E/Rt

I = 12/(24/17)

I = 12×17/24

I = 8.5A

If the connection is in parallel, the total current in the circuit will be 8.5A

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3 years ago

Read 2 more answers

There are several forms of energy. Which of the following is not a form of energy?

Evgen [1.6K]

A. Is the right answer.

8 0

3 years ago

Read 2 more answers

What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a

andrezito [222]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

6 0

4 years ago

A skier accelerates down the hill at 3m/s2 how fast is he going in 4 seconds​

A skier accelerates down the hill at 3m/s2 how fast is he going in 4 seconds