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FinnZ [79.3K]
3 years ago
11

A bear fells from 10m long grass From where the bear is

Physics
1 answer:
Elina [12.6K]3 years ago
3 0
He from the 10m long grass
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A car is accelerated from 40 m/s to 48 m/s in 6.5 seconds. What is the magnitude of the car’s
pantera1 [17]

Answer:

Explanation:

AVerage acceleration is the cjange in velocity with time

a = v-u/t

v is the final velocity = 48m/s

u is the initial velocity = 40m/s

t is the time = 6.5s

a = 48-40/6.5

a = 8/6.5

a = 1.23m/s²

Hence the magnitude of the car’s  average acceleration during this period is 1.23m/s²

4 0
3 years ago
Use the drop-down menus to identify the parts of DNA. Label A: Label B:
love history [14]

Label A:

Nitrogen bases

Label B:

Sugar-phosphate backbone

7 0
3 years ago
Read 2 more answers
Which one of the following is NOT a simple machine?
ivann1987 [24]
It is hammer because hammers are not examples of a simple machine
6 0
3 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
The car in the figure travels at a constant speed along the road shown. Draw vector showing its acceleration at the point C if t
yuradex [85]

Answer:

The vector form is as shown in the attachment

Explanation:

The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.

Mathematically from centripetal acceleration; a = v2/r

The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.

The attachment shows the figure and the representation in vectorial form.

6 0
3 years ago
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