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Softa [21]
3 years ago
11

An ant sits on the back of a mouse. The mouse carries the ant across the floor for a distance of 150m to her bedroom. How much w

ork did the mouse do?
Physics
1 answer:
marin [14]3 years ago
8 0

Answer:

Since the weight of the ant is not given, we cannot determined the work done.

Explanation:

Given parameters:

Distance covered  = 150m

Unknown:

Work done by mouse  = ?

Solution:

To solve this problem, we need to understand that work done is the force applied to move a body through a certain distance.

In this case, work done;

    Work done  = force x distance

     Work done  = Weight x distance

Since the weight of the ant is not given, we cannot determined the work done.

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A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total
NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is 64.59^{\circ} South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is 64.59^{\circ} South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, \vec{AB} = 446 km

Distance moved from B to C in South direction, \vec{BC} = - 939 km

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

AC = \sqrt{(AB)^{2} + (BC)^{2}}

AC = \sqrt{(446)^{2} + (- 939)^{2}} = 1039 km

(b) Direction of the displacement is given by:

tan\theta = \frac{\vec{BC}}{\vec{AB}}

\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}

64.59^{\circ} South of East

(c) Magnitude of the average speed is given by:

v_{avg} = \frac{AC}{t + t'}

v_{avg} = \frac{1039}{1.82} = 570.88 km/h

(d) The direction of the average velocity is the same as that of the displacement, i.e., 64.59^{\circ} South of East.

(e) The average speed of the [plane is given by:

v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}

v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h

6 0
3 years ago
The moment of inertia of a thin ring of mass M and radius R about its symmetry axis is ICM = MR2.
monitta

Answer:

The moment of inertia of large ring is 2MR².

(A) is correct option.

Explanation:

Given that,

Mass of ring = M

Radius of ring = R

Moment of inertia of a thin ring = MR²

Moment of inertia :

Moment of inertia is the product of the mass of the ring and square of radius of the ring.

We need to calculate the moment of inertia of large ring

Using formula of moment of inertia

I=I_{cm}+MR^2

Where, I_{cm} = moment of inertia at center of mass

M = mass of ring

R = radius of ring

Put the value into the formula

I=MR^2+MR^2

I=2MR^2

Hence, The moment of inertia of large ring is 2MR².

6 0
3 years ago
A stunt woman of mass m falls into a net during the filming of an action movie. Assume she experiences upward acceleration magni
Rashid [163]

Answer:

M= F^n / a+g

Explanation:

This shows correctly Newton’s second law, where sum of forces is divided by mass is equal to acceleration. Also mass can’t be negative so F^n is positive.

8 0
3 years ago
Read 2 more answers
1.5-m length of straight wire experiences a maximum force of 1.2 N when in a uniform magnetic field that is 1.8 T. 1) What curre
Iteru [2.4K]

Answer:

 I = 0.44 A

Explanation:

The magnetic force on a conductor is given by the expression

       F = I L x B

Where bold letters indicate vectors, I is the current, L is the vector in the direction of the current, and B is the magnetic field

Since the force is maximum, the wire must be perpendicular to the magnetic field, therefore

        F = I L B sin 90

        I = F / L B

Let's calculate

        I = 1.2 / 1.5 1.8

        I = 0.44 A

7 0
3 years ago
Read 2 more answers
Cells are known as the foundation of all life.How have these small structures changed our planet?
ELEN [110]
They are the very bases of life, no life would exist without cells
5 0
3 years ago
Read 2 more answers
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