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timofeeve [1]
3 years ago
11

6. If a vehicle needs 5s to complete 15m. Find the mean speed of it?

Physics
1 answer:
Karolina [17]3 years ago
4 0

Answer:

3m/s

Explanation:

Time=5s

Distance =15m

Speed=distance/time

Putting the values

Speed=15m/5s

Speed=3m/s is the answer

Hope it will help you. :)

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What does gravitational potential energy depend on?
jolli1 [7]
It depends on the mass of an object and acceleration because of the gravity and the height of an object
3 0
3 years ago
A red car passes a blue car. Which is true?
Marrrta [24]
D. The red car is moving faster than the blue car
4 0
3 years ago
An 3.7 lb hammer head, traveling at 5.8 ft/s strikes a nail and is brought to a stop in 0.00068 s. The acceleration of gravity i
CaHeK987 [17]

Answer:

31677.2 lb

Explanation:

mass of hammer (m) = 3.7 lb

initial velocity (u) = 5.8 ft/s

final velocity (v) = 0

time (t) = 0.00068 s

acceleration due to gravity (g) 32 ft/s^{2}

force = m x ( a + g )

where

  • m is the mass = 3.7 lb
  • g is the acceleration due to gravity = 32 ft/s^{2}
  • a is the acceleration of the hammer

       from v = u + at

       a = (v-u)/ t

       a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)

we can substitute all required values into force= m x (a+g)

force = 3.7 x (8529.4 + 32) = 31677.2 lb

       

4 0
3 years ago
The experimentation step of scientific inquiry involves _______. A. Analyzing data b. Drawing a conclusion c. Choosing variables
Anna11 [10]

Answer:

C- Choosing variables and controls

Explanation:

Correct on edge.

3 0
2 years ago
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
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