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Varvara68 [4.7K]
3 years ago
9

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk

. After 14sec, how far will it have traveled in the direction parallel to the edge of the sidewalk?
A) 0.005m
B) 0.01m
C) 1.0m
D) 9.9m​
Physics
2 answers:
ahrayia [7]3 years ago
6 0

Answer:

C) 1.0 m

Explanation:

The component of the velocity parallel to the sidewalk is:

vₓ = v cos θ

vₓ = 0.1 m/s cos 45°

vₓ = 0.0707 m/s

The distance traveled after 14 seconds is:

d = vₓ t

d = (0.0707 m/s) (14 s)

d = 0.99 m

Closest answer is C) 1.0 m.

DerKrebs [107]3 years ago
5 0

Answer:

C) 1.0m

Explanation:

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk. After 14sec, it will have traveled 1.0m in the direction parallel to the edge of the sidewalk.

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Answer:

200000 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

KE => is the kinetic energy.

m =>is the mass of the object

V => it the velocity of the object.

With the above formula, we can obtain the kinetic energy of the roller coaster as follow:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 1000 × 20²

KE = 500 × 400

KE = 200000 J

Therefore, the kinetic energy of the roller coaster is 200000 J.

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The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t
Salsk061 [2.6K]

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2}  \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\

Now, we calculate the strain:

strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain =  0.002\\

Now, we will calculate the Young's Modulus (Y):

Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\

<u>Y = 4.775 x 10⁹ Pa = 4.775 GPa</u>

6 0
3 years ago
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