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2 years ago

A tourist averaged 82km/hr for a 6.5h trip in her volkswagen. how far did she go?

1 answer:

4 0

82 ÷ 6.5 = 12.615384615384... repeating
round = 12.6
12.6 · 6.5 = 81.9
round = 82
She went an average of 12.6 km an hour
Hope this helps! ;)

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An object in equilibrium has a net force of

Butoxors [25]

Answer:

An object in equilibrium has a net force of zero

Static equilibrium describes an object at rest having equal and balanced forces acting upon it.

Dynamic equilibrium describes an object in motion having equal and balanced forces acting upon it.

Explanation:

An object is said to be in equilibrium when a net force of zero is acting on it. When this condition occurs, the object will have zero acceleration, according to Newton's second law:

$F=ma$

where F is the net force, m the mass of the object, a the acceleration. Since F=0, then a=0. As a result, we have two possible situations:

- If the object was at rest, then it will keep its state of rest. In this case, we talk about static equilibrium.

- If the object was moving, it will keep moving with constant velocity. In this case, we talk about dynamic equilibrium.

8 0

2 years ago

Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

DiKsa [7]

Am not sure
Have a nice day

6 0

2 years ago

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

6 0

2 years ago

In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one sid

mojhsa [17]

Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

5 pull 62 N each to the left

3 pull 150 N each to the right

1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

$Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110$

The resulting force is negative, meaning the direction is to the left

6 0

2 years ago

A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th

Stolb23 [73]

Answer:

The current is $I_b = 400 \ A$

Explanation:

From the question we are told that

The length of the segment is $l = 2.50 \ m$

The current is $I_a = 1000 \ A$

The force felt is $F = 4.0 \ N$

The distance of the second wire is $d = 5.0 \ cm = 0.05 \ m$

Generally the current on the second wire is mathematically represented as

$I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4 \pi * 10^{-7} \ N/A^2$

=> $I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }$

=> $I_b = 400 \ A$

4 0

2 years ago