Ask question

Ask question

All categories

alexandr1967 [171]

3 years ago

11

Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0

.7 km in the original direction. Let Tiana's original direction be the positive direction. What are the displacement and distance she jogged?

1 answer:

vichka [17]3 years ago

7 0

Answer:

Distance: 4.6km Displacement= -0.2km

Explanation:

Total distance: 1.5+2.4+0.7= 4.6 km

Displacement: 1.5-2.4+0.7= -0.2km

The displacement may also be 0.2km, it just depends on if it wants it negative or not.

You might be interested in

What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a

Nutka1998 [239]

The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:

Vf = Vi + at

where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time

Substituting the values gives:

30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>

6 0

2 years ago

Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared

Aloiza [94]

Answer:

$Volume = 6248.48 m^{3}$

Explanation:

Given:

The area of the house $A = 2050\ ft^{2}$

The height of the house $h=10\ ft$

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

$Volume = Area\times height$

$Volume = A\times h$

Area and height of the house are known, so we substitute these value in above equation.

$Volume = 2050\times 10$

$Volume = 20500\ ft^{3}$

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for $m^{3}$

$Volume = \frac{20500}{3.2808}$

$Volume = 6248.48\ m^{3}$

Therefore, the volume of the house is $6248.48 m^{3}$

8 0

3 years ago

A player at first base catches a throw traveling 38 m/s. The baseball, which has a mass of 0.145 kg, comes to a complete stop in

mixas84 [53]

Answer:

F = 39.36 N

Explanation:

given,

initial speed, u = 38 m/s

final speed, v = 0 m/s

mass of ball = 0.145 Kg

time, t = 0.14 s

Force = ?

using impulse formula

J = change in momentum

J = F x t

m(v - u) = F x t

0.145 x (0 - (-38)) = F x 0.14

F x 0.14 = 5.51

F = 39.36 N

force exerted by the ball is equal to 39.36 N.

5 0

3 years ago

What is the magnification of a real image if the image is 15.0 cm from a lens and the object is 60.0 cm from the lens?

Hoochie [10]

Answer:

soluble

Explanation:

7 0

2 years ago

Read 2 more answers

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

$m' = 105Kg$

4 0

3 years ago