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2 years ago

What is the velocity of a car if it travels east 340 meters in 10 seconds? V = d/t

Physics

2 answers:

Nadya [2.5K]2 years ago

6 0

Answer:

The awnser is d

Explanation:

i know cause i took the test

erastovalidia [21]2 years ago

5 0

Answer:

D 34 m/s east

Explanation:

Took the test got 100%

You might be interested in

A magnetic force can act on an electron even when it A) is at rest B) moves parallel to magnetic field lines C) both of these D)

Kobotan [32]

Answer: A)

Explanation: when an electron is placed in a magnetic field, it experiences a force.

This force is given below as

F=qvB*sinθ

F = force experienced by charge.

q = magnitude of electronic charge

v = speed of electron

B= strength of magnetic field

θ = angle between magnetic field and velocity.

What defines the force exerted on the charge is the angle between the field and it velocity.

If magnetic field is parallel to velocity, then it means that θ=0° which means sin 0 = 0, which means

F = qvB * 0 = 0.

The charge being at rest has nothing to do with the angle between magnetic field strength and velocity.

3 0

3 years ago

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

m1 = 0.0225 Kg

V1 = (0.869, -0.283) m / s

Px = m Vx

Px1 = 0.0225 0.869

Px1 = 0.01955 Kg m

Py = m Vy

Py1 = 0.0225 (-0.283)

Py1 = -0.006368 Kg m

P1 = (0.0196, -0.00637) Kg m

Mouse 2

m2 = 0.0223 Kg

Px2 = 0.0223 (-0.883) = -0.0196 Kg m

Py2 = 0.0223 (-0.253) = -0.00564 Kg m

P2 = (-0.0196, -0.00564) Kg m

Mouse 3

m3 = 0.0197

Px3 = 0.0197 0.345 = 0.00680 Kg m

Py3 = 0.0197 0.803 = 0.0158 Kg m

P3 = (0.00680, 0.0158) Kg m

Mouse4

m4 = 0.0127 Kg

Px4 = 0.0127 (-0.555) = -0.00705 Kg m

Py4 = 0.0127 0.205 = 0.00260 Kg m

P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

Px = Px1 + Px2 + Px3 + Px4

Py = py1 + Py2 + Py3 + Py4

Px = 0.0196 -0.0196 +0.00680 -0.00705

Px = -0,000250 Kg m

Py = -0.00637 -0.00564 +0.0158 +0.00260

Py = 0.00639 Kg m

P = (-0.000250, 0.00639) Kg m

7 0

3 years ago

A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a

zloy xaker [14]

Answer:

v = 1.4 m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5 x v² = 9.8 x .1

= v² = 1.96

v = 1.4

v = 1.4 m /s

6 0

2 years ago

Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth,

Alex17521 [72]

Answer:

A) 1.67 x 10 ⁻⁶ m/s

B)5.59 x $10^-^9$ %

Explanation:

Given:

d = 5.0 km,

mₐ = 2.5 x $10^1^4$ kg

u₁ = 4.0 x 10⁴ m/s

$m_n$ = 5.98 x 10 ²⁴ kg

Solve using kinetic conserved energy

mₐ x u₁ + $m_n$ x u₂ = uₓ x (mₐ + $m_n$ )

(2.5 x $10^1^4$ ) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x $10^1^4$ + 5.98 x 10 ²⁴ )

uₓ = ( 2.5 x $10^1^4$ x 4.0 x 10⁴ ) / (2.5 x $10^1^4$ + 5.98 x 10 ²⁴ )

uₓ = 1.67 x 10 ⁻⁶ m/s

B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m

t = 365 days x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s

t = 31536000 s

D = 2 π R = 2 π( 1.5 x 10 ¹¹ )

D = 9.4247 x 10 ¹¹ m

u₂ = D / t = 9.4247 x 10 ¹¹ / 31536000

u₂ = 29885.775 m/s

% = ( 1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100

% = 5.59 x $10^-^9$ %

5 0

3 years ago

Object A attracts object B with a gravitational force of 10 newtons from a given distance. If the distance between the two objec

balandron [24]

By definition, the gravitational force is given by:

$F = G(\frac{m1m2}{r^2})$

Where,

G: gravitational constant

m1: mass of object 1

m2: mass of object 2

r: distance between objects:

We know that the force is equal to 10N:

$G(\frac{m1m2}{r^2}) = 10$

If the distance is double we have:

$F = G(\frac{m1m2}{(2r)^2}) = G(\frac{m1m2}{4r^2}) = \frac{1}{4}G(\frac{m1m2}{r^2})$

$\frac{1}{4}(10) = 2.5$

Therefore, the new force is:

2.5 newtons

Answer:

the changed force of attraction between them is:

A. 2.5 newtons

7 0

3 years ago

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