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dusya [7]
3 years ago
7

The average speed of a car being driven in London is 11 miles per hour. If you have to drive 15 miles from your home to work in

London how long do you expect it to take
Physics
1 answer:
arlik [135]3 years ago
4 0
It would take you approximately  an hour and 15 mins
You might be interested in
22. A ball is thrown horizontally from the roof of a building 12 m tall with a speed of 3.1 m/s.
zysi [14]

Answer:

a) t = 1.6 s

b) d = 4.9 m

c) v = 16 m/s

d) θ = 79°

Explanation:

time of fall

t = √(2h/g) = √(2(12)/9.8) = 1.5649... s

d = vt = 3.1(1.56) = 4.8512...

vertical velocity vy = at = 9.8(1.56) = 15.336... m/s

v = √(15.336² + 3.1²) = 15.6464... m/s

θ = arctan(15.336/3.1) = 78.5724...°

5 0
3 years ago
On another planet gravity has a value of 5.5 m/s . If an object is dropped how long will it take to fall 53 m?
Nikolay [14]

Answer:

4.4 seconds

Explanation:

Given:

a = -5.5 m/s²

v₀ = 0 m/s

y₀ = 53 m

y = 0 m

Find: t

y = y₀ + v₀ t + ½ at²

0 = 53 + 0 + ½ (-5.5) t²

0 = 53 − 2.75 t²

t = 4.39

Rounded to two significant figures, it takes 4.4 seconds for the object to land.

7 0
3 years ago
Which of the following statements is TRUE about impulse? (Consider the Impulse-Momentum Theorem: F▲t = m ▲v )
kotykmax [81]

The statement “Impulse is a vector quantity” is true about Impulse.

Answer: Option B

<u>Explanation: </u>

The object’s action by applied force in a particular time interval, there happens changing in momentum called impulse. It is denoted by a symbol ‘J’ or ‘imp’ and expressed in a unit ‘Ns’. As impulse depends on the acted force, when a collision arises from front, behind or side, the force’s direction would be differed.

                   \text {Impulse }=\text {Force } \times \text {time}=\vec{F} \Delta t

So, from this option A is false as impulse is not a force but changing momentum. The unit is not Newton, it is Newton second (Ns). The force direction differs (impulse direction) for each cases of collision, so option D also false. Hence, option B seems to be correct. Vector quantity deals with both direction and magnitude and important in motion study.

8 0
3 years ago
Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.
ASHA 777 [7]

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

  • Tension is also a force having Newton as S.I unit.
  • The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using  Lami's theorem.

The sum of two given angles  = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}

Where

\alpha  = \beta = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

\frac{T}{sin\alpha } =  \frac{W}{sinY}

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

3 0
2 years ago
7. A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a level (frictionless) surfa
makkiz [27]

Answer:

0.505 m

Explanation:

From the question,

The kinetic energy of the car = energy stored in the spring

1/2mv² = 1/2ke²...................... Equation 1

Where m = mass of the car, v = velocity of the car, k = spring constant of the car, e = extension/compression

make e the subject of the equation

e = v√(m/k)............... Equation 2

We can calculate the value of v, by applying,

v² = u²+2gH...................... Equation 3

Where u = initial velocity of the car, H = height of the car, g = acceleration due to gravity.

Given: u = 0 m/s (from rest), H, 10 m, g = 9.8 m/s²

Substitute into equation 2

v² = 2(10×9.8)

v² = 196

v = √196

v = 14 m/s

Also given: m = 1300 kg, e = 1.0×10⁶ N/m =1000000 N/m

Substitute into equation 2

e = 14√(1300/1000000)

e = 14√(0.00013)

e = 14(0.036)

e = 0.505 m

Hence the maximum distance of the spring is compressed = 0.505 m

5 0
3 years ago
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