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Varvara68 [4.7K]

2 years ago

9

if an object is moving with constant speed then its distance time graph will be a straight line. a) along time axis b) along dis

tance axis c) parallel to time axis d)inclined to time axis.

1 answer:

Varvara68 [4.7K]2 years ago

5 0

Answer: I’m not sure but the time and distance will be increasing while they required the distance time so my answer is ((A))

Explanation:

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Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the b

RSB [31]

First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds

4 0

3 years ago

Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$ F be the temperature and T that of the body

Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = <em>kdt</em>

In [T-70]= <em>kt </em>+ $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for <em>k </em>using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

<em>K </em>= $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230 $e^{-0.19018}$

4 0

2 years ago

The total mechanical energy of a basketball is 400 J. If the kinetic energy is 286 J, what must the potential energy be?

ch4aika [34]

Answer:

the potential energy is 114 J.

Explanation:

Given;

total mechanical energy, E = 400 J

kinetic energy, K.E = 286 J

The potential energy is calculated as follows;

E = K.E + P.E

where;

P.E is the potential energy

P.E = E - K.E

P.E = 400 J - 286 J

P.E = 114 J

Therefore, the potential energy is 114 J.

7 0

2 years ago

How does electricity affect objects that are not in contact with one another?

Elan Coil [88]

Electrostatic forces are non-contact forces; they pull or push on objects without touching them

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1 year ago

A uniform brick of length 21 m is placed over

Paha777 [63]

Answer:

15.75 m

Explanation:

First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.

Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

6 0

3 years ago